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WINSLO W'S 



COMPREHENSIVE MATHEMATICS: 



K<\ 



BEING AN EXTENSIVE CABINET OF 



NUMERICAL, ARITHMETICAL, AND MATHEMATICAL FACTS, 
TABLES, DATA, FORMULAS, AND PRACTICAL RULES FOR 
THE GENERAL BUSINESS-MAN, MERCHANT, MECHANIC, 
ACCOUNTANT, TEACHERS OF SCHOOLS, GEOME- 
TRICIANS, AND SCIENTISTS; APPROPRIATELY 
ARRANGED AND APPLIED. 



BY 

E. S. WIN SL O W. 



Eighth Edition. Enlarged and Improved. 




BOSTON: 

PUBLISHED BY THE AUTHOR. 

1870. 



Single copies of Winslow's Mathematical Works are delivered at the coun- 
ter, or sent by mail, post-paid, on receipt of price, as follows, viz. : — 

Comprehensive Mathematics $2.50 

The Universal Modern Cambyst, and Foreign and Domestic 

Commercial Calculator 1.50 

The Tin-Plate and Sheet-Iron Workers' Monitor . . . 1.00 

Sold by G-. W. Lindsey, Bookseller and Stationer, No. 1031 Washington St., 
Boston. 



Entered, according to Act of Congress, in the year 1867, by E. S. Winslow, in 
the Clerk's Office of the District Court of the District of Massachusetts. 

Entered, according to Act of Congress, in the year 1870, by E. S. Winslow, 
in the Clerk's Office of the District Court of the District of Massachusetts. 



Stereotyped by C. J. Peters & Son, 
5 Washington Street. 



PREFACE 
TO THE COMPKEHENSIVE MATHEMATICS. 



On presenting this work to the public, it may be proper to state 
that it has been designed and written mainly for the practical 
man. It contains a vast array of Numerical, Arithmetical, and 
Mathematical facts, tables, data, formulas, and rules, pertaining 
to a great variety of subjects, and applicable to a diversity of ends, 
as well as much information of a more general nature, valuable to 
the artisan, and commercial classes ; thus meeting the wants, in an 
eminent degree, of the lovers of the exact sciences, and the prac- 
tical wants of students in the mathematics. 

The facts and data alluded to have been gathered, with much 
care and patience, from a great variety of sources, or derived, often 
by toilsome investigations, from known and accredited truths. 
The care that has been taken in respect to* these, it is thought, 
should secure for this particular department reliance and trust. 

The tables, which are numerous, have, with few exceptions, been 
composed and arranged expressly for the work, and a confidence 
is felt that they may be relied on for accuracy. 

From the valuable works of Dr. Ure, Adcock, Gregory, Grier, 
Brunton ; from the publications of the transactions of London, 
Edinburgh, and Dublin Philosophical Societies ; and from the pub- 
lications by the Smithsonian Institute, much valuable information 
has been gained, relating mainly to machinery and the arts ; and 
to these sources the author feels indebted. 

The conciseness with which the work has been generally written 
would, perhaps, be found an objection, were it not that all the pro- 
positions and problems of intricacy are accompanied with exam- 
ples and illustrations, and, in the matters of Geometry, additionally 
accompanied with diagrams. The whole, it is thought, will appear 
clear to him who consults it. A prominent feature in the design 
has been to produce a useful work, and one which in the way ot 
price shall be readily accessible to all. 



iv PREFACE. 

PREFACE 

TO THE UNIVERSAL MODEM CAMBYST, AND FOREIGN 
AND DOMESTIC COMMERCIAL CALCULATOR. 



This work is composed of the first five sections of the author's 
" Comprehensive Mathematics." It was thought advisable 
to publish this portion of that work in a separate form on account 
of price ; more especially as it contains all of a commercial nature 
treated of in that work. Indeed, the contents of that work were 
arranged expressly to this end. The Table of Contents in both 
works is the same : the work being stereotyped, this could not 
well be avoided. The Table of Contents, therefore, in either 
work, is that of the " Comprehensive Mathematics ; " and 
the first five sections thereof, that is, Section I., Section II., Sec- 
tion III., Section A., and Section B., is that of the " Universal 
Modern Cambyst, and Foreign and Domestic Commercial Calcu- 
lator." 



PREFA C E 
TO THE TIN-PLATE AND SHEET-IRON WORKERS' MONITOR. 



This work is composed of Section VI. of the author's " Com- 
prehensive Mathematics," with portions of other sections of 
that work. It embraces all that is contained in the last-mentioned 
work of special interest to the Tinsmith, as such. It may be re- 
lied on for accuracy in all particulars, and is believed to be the 
first and only reliable work of the kind ever published. It is pub- 
lished in separate form on account of price, and with the view of 
affording apprentices and students every possible facility of obtain- 
ing it. It contains over 100 pages, nearly 50 diagrams, and step- 
by-step directions for constructing, mechanically, not less than 30 
unlike and different patterns, embracing all of the more difficult 
and complicated in use, and several of new and beautiful designs. 



CONTENTS. 



SECTION A. 

PAGE 

Foreign Moneys of Account ... al 
Foreign Linear and Surface Meas- 
ures . «18 

Foreign Weights «26 

Foreign Liquid Measures .... «37 

Foreign Dry Measures a44 

Memoranda, &c., relative to Foreign 

Moneys, &c. * all «50 

United-States Customs' Tares . . abl 



SECTION I. 



moneys of account, coins, 
weights, and measures op 
the united states ; foreign 
gold coins, &c. 

Explanations of Signs .... 12 

Moneys of Account of the United 

States 13 

Comparative Yalue of Gold and 

Silver 13 

Gold, pure ; value of, by weight . . 15 
Mint Gold, Standard of, &c. . . . 15 
Gold Coins, their weights and val- 
ues 15 

Silver, pure; value of, by weight . 16 
Mint Silver, Standard of, &c. . . . 16 
Silver Coins, their weights and 

values 16 

Copper Coins, &c 16 

Present Par Value of Silver Coins 

issued prior to June, 1853 .... 17 
Currencies of the different States 

of the Union 17 

The Metrical System of Weights 

and Measures 18 

Foreign Gold Coins, Tables of, &c. 19 
Foreign Silver Coins, Values of, 25 

1* 



WEIGHTS AND MEASURES. 

PAGB 

Long or Linear Measure ... 25 

Cloth Measure 25 

Land Measure 25 

Engineer's Chain 25 

Shoemaker's Measure 26 

Miscellaneous Measures 26 

Square or Superficial Meas- 
ure 26 

Measure for Land . ,26 

Circular Measure 27 

Cubic or Solid Measure ... 27 
General Measure of Weight, 28 

Gross Weight 28 

Troy Weight 28 

Apothecaries' Weight, 28 

Diamonds, Measure of Value, &c, 28 

Liquid Measure 28 

Imperial Liquid Measure 29 

Ale Measure 29 

Dry Measure . . . 29 

Imperial Dry Measure 30 



SECTION II. 



miscellaneous facts, calcu- 
lations, AND MATHEMATICAL 
DATA. 

Specific Gravities, Tables of, 31 

Weight per Bushel of Articles . . 35 
Weight per Barrel of Articles . . . 35 
Weights of different Measures of 

various Articles 35 

Weight of Coals, &c, Tables . 35, 55 
Practical Approximate Weight in 

Pounds of Various Articles ... 36 



Ropes and Cables 
v 



CONTENTS. 



PAGE 

Weight and Strength of Iron 
Cnains 37 

Comparative Weight of Metals, 
Table 38 

Weight of Rolled Iron, Square Bar, 
Tables 38 

Weight of Various Metals, differ- 
ent Forms of Bar 39 

Weight of Bound-rolled Iron, Ta- 
ble 40 

Weight of Cast-iron Prisms of dif- 
ferent forms, &c 40 

Weight of Flat-rolled Iron, Table, 42 

Weight of Different Metals,in Plate, 44 

The American Wire Gauge . 45 

The Values of the Nos. American 
Wire Gauge and Birmingham 
Wire Gauge, in the United States, 

inch, Tables of 45 

The Number of Linear Feet in a 
Pound of different kinds of Wire 
of different Sizes, Table of, &c, 46 
Characteristics, &c, of Alloys of 

Copper and Zinc,— BRASS . ... 47 
The Weight per Square Foot of dif- 
ferent Rolled Metals of different 
thicknesses by the Wire Gauge, 

Table 48 

Tin Plates, Sizes, &c, Table . 49 
Sheet Iron, Sheet Zinc, Copper 
Sheathing, Yellow Metal, Weight 

of, &c 49 

Capacity in Gallons of Cylindrical 

Cans, &c, Table 50 

Weight of Pipes 52 

Weight of Pipes, Table 53 

Weight of Cast-iron and Lead 

Balls 54 

Weight of Hollow Balls or Shells, 54 

Analysis of Coals 55 

Weight, Heating Power, &c, of 
Coals and other kinds of Fuel, 
Table 55 

Mensuration of Lumber ... 56 

Board Measure 56 

To Measure Square Timber .... 56 

To Measure Round Timber .... 56 

Table relative to the Measurement 
of Round Timber 57 

to find the Solidity of the greatest 
Rectangular Stick that can be cut 
from a Log 1 of Given Dimensions, 58 

To find the Solidity of the greatest 
Square Stick that can be cut from 
a Round Stick of Given Dimen- 
sions . 59 

To find the Contents of a Log in 
Board Measure 59 

Gauging 60 



PAGE 

To find the Dimensions of Vessels 
of different Forms, for holding 
Given Quantities 62 

Cask Gauging, all Forms of 
Casks 63 

To find the Contents of a Cask, the 
same as would be given by the 
Gauging Rod 66 

To find the Diagonal and Length 
of a Cask 66 

Ullage 67 

To find the Ullage of a Standing 
Cask 67 

To find the Ullage when the Cask is 
upon its Bilge 67 

To find the Quantity of Liquor in a 
Cask by its Weight 68 

Customary Rule by Freighting Mer- 
chants for finding the Cubic 
Measurement of Casks 68 

Tonnage of Vessels, to Calcu- 
late 69 

Of Conduits, or Pipes 70 

To find the requisite thickness of a 
Pipe to support a Given Head of 
Water 70 

To find the Velocity of Water pass- 
ing through a Pipe 71 

To find the Head of Water requi- 
site to a Required Velocity 
through a Pipe 71 

To find the Quantity of Water Dis- 
charged by a Pipe in a Given 
Time 71 

To find the Specific Gravity of a 
Body heavier than Water .... 72 

To find the Specific Gravity of a 
Body lighter than Water .... 72 

To find the Specific Gravity of a 
Fluid 72 

To find the Quantity of each of the 
several Metals composing an Al- 
loy 72 

To find the Lifting-power of a Bal- 
loon 73 

To find the Diameter of a Balloon 
equal to the Raising of a Given 
Weight .....' 73 

To find the Thickness of a Hollow 
Metallic Globe that shall have a 
Given Buoyancy in a Given 
Liquid 72 

To Cut a Square Sheet of Metal so 
as to form a Vessel of the Great- 
est Capacity the Sheet admits of . 73 

Comparative Cohesive Forces of 
Substances, Table ........ 74 

Alloys having a Tenacity greater 
than the Sum of their Con- 
stituents 74 



CONTENTS. 



PAGE 

Alloys having a Density greater 
than the Mean of their Con- 
stituents 75 

Alloys having a Density less than 

the Mean of their Constituents . 75 
Relative Powers of different Metals 

to Conduct Electricity 75 

Dilations of Solids by Heat, Table 75 
Melting Points of Metals and other 

Substances, Table 76 

Relative Powers of Substances to 

Radiate Heat, Table 76 

Boiling Points of Fluids 76 

Freezing Points of Fluids .... 77 
Expansion of Fluids by Heat . . . 77 
Relative Powers of Substances to 

Conduct Heat 77 

Ductility and Maleability of Metals, 77 
Quantity per cent, of Nutritious 
Matter contained in different Ar- 
ticles of Food 78 

Standard, &c, of Alcohol 78 

Quantity per cent, of Absolute Al- 
cohol contained in different Pure 
Liquors, Wines, &c, Table . . 78 
Proof of Spirituous Liquors ... 78 
Comparative Weight of Timber in 
a Green and Seasoned State, Ta- 
ble, &c 79 

Relative Power of different kinds of 

Fuel to Produce Heat, Table, . 79 
Relative Illuminating Power of dif- 
ferent Materials, Table and Re- 
marks, 80 

Thermometers, different kinds, 
to Reduce one to another, &c, . 82 

Horse- Power. 83 

Animal Power 83 

Steam, Tables in relation to, 

&c, 83, 308 

Velocity and Force of Wind, Ta- 

•ble 84 

Curvature of the Earth . . . . 84, 213 
Degrees of Longitude, Lengths of, 

&c, 84 

Time, with respect to Longitude, 84 

Velocity of Sound 84 

Velocity of Light 85 

Gravitation ........ 85, 302 

Area of the Earth, its Density, &c, 85 

Cnemical Elements 86 

Elementary Constituents of Bodies, 

Table 87 

Combinations by Weight of the 
Gases in forming Compounds, 

Table 87 

Combinations by Volume of the 
Gases, their Condensation, &c, 

in forming Compounds 89 

Atomic Weight . , 89 



Chemical and other Properties of 
Various Substances 90 



SECTION III. 



practical arithmetic. 

Vulgar Fractions 95 

Reduction of Vulgar Fractions . . 95 
Addition of Vulgar Fractions ... 99 
Subtraction of Vulgar Fractions . 99 
Division of Vulgar Fractions . . . 100 
Multiplication of Vulgar Fractions 100 
Multiplication and Division of 

Fractions Combined 101 

Cancellation 96, 97, 102 

To Reduce a Fraction in a higher, 
to an equivalent in a given low- 
er denomination 102 

To Reduce a Fraction in a lower, 
to an equivalent in a given high- 
er denomination 102 

To Reduce a Fraction to Whole 
Numbers in lower given denom- 
inations .... 103 

To Reduce Fractions in lower de- 
nominations to given higher de- 
nominations 103 

To work Vulgar Fractions by the 

Rule of Three, or Proportion . . 104 
Decimal Fractions ...... 104 

Addition of Decimals 105 

Subtraction of Decimals 105 

Multiplication of Decimals .... 106 

Division of Decimals 106 

Reduction of Decimals 107 

To work Decimals by the Rule of 

Three 108 

Proportion, or Rule of Three ... 109 

Compound Proportion 110 

Conjoined Proportion, or Chain 

Rule ..112 

Percentage 114 

Interest 120 

Compound Interest 122 

Bank Interest, or Bank Discount . 127 

Discount 129 

Compound Discount 129 

Profit and Loss . 130 

Equation of Payments. 132 

General Average 134 

Assessment of Taxes 136 

Insurance 136 

Life Insurance • 136 

Fellowship 138 



CONTENTS. 



PAGB 

Alligation . 139 

Involution 141 

Evolution 141 

To Extract the Square Root ... 142 
To Extract the Cube Root .... 143 

To Extract any Root 145 

Arithmetical Progression 146 

Geometrical Progression 150 

Annuities 154 

Of Installments generally .... 164 

Permutation 166 

Combination 167 

Problems 169 



SECTION IV. 

geometry. 

Definitions, Construction of 

Figures, &c 172 

To Bisect a Line 176 

To Erect a Perpendicular 176 

To Let Fall a Perpendicular ... 176 

To Erect a Perpendicular on the 
end of a Line 177 

To draw a Circle through any three 
points not in a straight line, and 
to find the Centre of a Circle, or 
Arc 177 

To find the Length of an Arc of a 
a Circle approximately by Me- 
chanics 177 

From a given Point to draw a 
Tangent to a Circle 177 

To draw from or to the Circumfer- 
ence of a Circle, lines tending 
to the Centre, when the latter is 
inaccessible 177 

To describe an Oval Arch on a 
given Conjugate Diameter ... 178 

To describe an Oval of a given 
Length and Breadth ...... 178 

To describe an Are or Segment of 
a Circle of Large Radius .... 179 

To describe an Oval Arch, the 
Span and Rise being given . , . 179 

Gothic Arches, to draw ...... 180 

Polygons, to construct ...... 181 

Polygons, to inscribe in a given 
Circle 181 

Polygons, to circumscribe about a 
given Circle , , .,181 

To produce a Square of the same 
Area as a given Triangle . . . .181 

To construct a Parabola . . .182, 355 

To Construct a Hyperbola . . 182, 349 

To bisect any given Triangle . . , 182 



To draw a Triangle equal in Area 
to two given Triangles^ . ... 183 

To describe a Circle equal in Area 
to two given Circles 183 

To construct a Tothed, or Cog- 
wheel 183 



Of the Conic Sections 



. 184 



Mensuration of Lines and Super- 
ficies. 

Triangles 185 

Of Right- Angled Triangles . . • 186 
Of Oblique- Angled Triangles . . 187 
To find the Area of a Triangle . 188 
To find the Hypotenuse of a Tri- 
angle 189 

To find the Base, or Perpendicu- 
lar, of a Triangle 188, 189 

To find the Height of an inacces- 
sible Object 189 

To find the Distance of an inac- 
cessible Object 190 

To find the Area of a Square, 
Rectangle, Rhombus, or Rhom- 
boid 190 

To find the Area of a Trapezoid . 191 
To find the Area of a Trapezium . 191 
Of Polygons, Table, &c. ... 194 
To find the Perpendicular of a 
Rhombus, Rhomboid, or Trape- 
zoid 192 

To find the Diagonal of a Rhom- 
bus, Rhomboid, or Trapezoid . 192 
To find the Area of a regular or 

irregular Polygon 195 

Circle , 196 

The Circle and its Sections .... 197 
To find the Diameter, Circumfer- 
ence, and Area of a Circle . . . 198 
To find the Length of an Arc of a 

Circle 1*99 

To find the Area of a Sector of a 

Circle . 201 

To find the Area of a Segment of 

a Circle 201 

To find the Area of a Zone ... 202 
To find the Diameter of a Cifcle of 

which a given Zone is a part . . 202 
To find the Area of a Crescent . . 202 
To find the Side of a Square that 
shall contain an Area equal to 

that of a given Circle 202 

To find the Diameter of a Circle 
that shall have an Area equal to 

that of a given Square 202 

To find the Diameters of three 
equal circles the greatest that 
can be inscribed in a given Cir- 
cle ....... . 202 



CONTENTS. 



PAGE 

To find the Diameters of four equal 
circles the greatest that can be 
inscribed in a given Circle . . . 202 
To find the Side of a Square in- 
scribed in a given Circle .... 203 
To find the Diameter of a Circle 
that will circumscribe a given 

Triangle 203 

To find the Diameter of the great- 
est Circle that can be inscribed 

in a given Triangle 203 

To divide a Circle into any num- 
ber of Concentric Circles of 

equal Areas 204 

To find the Area of the space con- 
tained between two Concentric 

Circles 205 

Ellipse 205 

To find the Area of an Ellipse . . 207 
To find the Length of the Circum- 
ference of an Ellipse 207 

To find the Area of an Elliptic Seg- 
ment 207 

Parabola 209 

To find the Area of a Parabola. . 210 
To find the Area of a Zone of a 

Parabola 210 

To find the Altitude of a Parabola, 210 
To find the Length of a Semi-para- 
bola 210 

Hyperbola 211 

To find the Length of a Semi- 
hyperbola 212 

To find the Area of a Hyperbola . 212 

Cycloid, and Epicycloid ... 212 

To find the Length of the Curve of 

a Cycloid 213 

To find the Area of a Cycloid. . . 213 
To find the Distance of Objects at 

Sea, &c 213 

Stereometry, or Mensuration 
of Solids. 

Of Prisms 214 

Of Right Prisms or Cubes .... 215 
Of Parallelopipedons .... 215 

Of Pyramids 215 

Of Frustums of Pyramids . . 216 

Of Prismoids 210 

Of the Wedge 217 

Of Cylinders 217 

To find the Length of a Helix . . 217 

Of Cones 218 

Of Frustums of Cones. . . .65, 218 
Of Spheres or Globes .... 219 



PAGE 

Of Spherical Segments 219 

Of Spherical Zones 220 

To find the greatest Cube that can 

be cut from a given Sphere . . . 220 

Of Spheroids 221 

Of Segments of Spheroids .... 221 
Of the Middle Frustum of a Spher- 
oid 65,221 

Of Spindles . ■ 222 

Of the Middle Frustum of a Para- 
bolic Spindle 65, 222 

Of Parabolic Conoids ... 65, 223 

Of Hyperboloids 223 

To find the Surface of a Cylindri- 
cal Ring 224 

To find the Solidity of a Cylindri- 
cal Ring 224 

Of the regular bodies .... 225 
Promiscuous Examples in 

Geometry . 226 

Trigonometry . 231 

Tables of Sines, Cosines, Tan- 
gents, &c 241 

Tables of Squares, Cubes, 

Square and Cube Roots, &c. 245 



SECTION V. 

MECHANICAL POWERS, MECHANI- 
CAL CENTRES, CIRCULAR MO- 
TION, STRENGTH OF MATERI- 
ALS; STEAM, THE STEAM EN- 
GINE, ETC. 

The Lever 271 

The Wheel and Axle .... 272 

The Pulley * . . . . 273 

The Inclined Plane 274 

The Wedge 275 

The Screw -275 

Transverse Strength of Bodies . 279 

Deflections of Shafts, &c 286 

Resistance of Bodies to Tortion . 287 
Resistance of Bodies to Compres- 
sion 289 

Centres of Surfaces 291 

Centres of Solids 293 

Centres of Oscillation and 

Percussion 294 

Centre of Gyration 298 

Central Forces 300 



CONTENTS. 



PAOB 

Fly Wheels 301 

The Governor 301 

Force of Gravity 302 

To find the Height of a Stream 

projected vertically from a Pipe, 303 
To find the Power requisite to 
project a Stream to any given 

Height 303 

Of Pendulums 304 

Screw-Cutting in a' Lathe . . 305 
Table of Change Wheels for 

Screw-Cutting in a Lathe . . . 308 
Of Steam and the Steam En- 
gine 308 

Velocity of Projectiles, &c . . . . 313 
Steam, acting expansively .... 313 
Of the Eccentric in a Steam En- 
gine 314 

Of Continuous Circular Mo- 
tion 314 

To find the number of Revolu- 
tions made by the last, to one 
revolution of the first, in a train 
of Wheels and Pinions .... 315 
The distance from Centre to Cen- 
tre of two Wheels to work in 
contact given, and the ratio of 
Velocity between them, to find 
their Requisite Diameters . . . 317 
To find the Velocity of a Belt . . 317 
To find the Draft on a Machine . 317 
To find the Revolutions of the 

Throstle Spindle 318 

To find the Twist given to the 

Yarn by the Throstle 318 

Teeth of Wheels, &c 318 

To construct a Tooth, &c 319 

To find the Horse-Power of a 

Tooth 319 

Journals of Shafts 320 

Hydrostatics 320 

Hydraulics 322 

Water- Wheels 323 

To find the Power of a Stream . .324 
To construct a Water- Wheel to a 

Given Power and Fall 325 

Dynamics 326 

Hydrostatic Press 326 



SECTION VI. 

coverings of solids, or prob- 
lems in pattern cutting. 

Remarks and Definitions . . . 333 



PAOB 

To construct a Pattern for the 
Lateral Portion of a vessel in the 
form of a Frustum of a Cone of 
given diameters and depth . . . 335 

To construct a Pattern for the 
Body of a vessel in the form of a 
Frustum of a Cone of given di- 
mensions, without plotting the 
dimensions 338 

To construct a Pattern for the 
Lateral Portion of a Flaring Ves- 
sel of given symmetry of outline 
and given capacity 339 

Table of Relative Propor- 
tions, Chords, &c 339 

The special tabular figure, the di- 
ameter of one end, and the Cubic 
Capacity of the vessel being 
given, to find the diameter of 
the other end 342 

To construct a Pattern for the body 
of a Flaring Vessel of .given 
tabular outline, and given dimen- 
sions, without plotting the di- 
mensions 344 

The Capacity in gallons of a vessel 
in the form of a Frustum of a 
Cone being given, and any two 
of its dimensions, to find the 
other dimension 346 

To construct Patterns for flaring 
oval vessels of different eccentri- 
cities and given dimensions, Nos. 
1,2,3 348 

To describe the bases for Nos. 1,2, 3, 349 

Of Cylindrical Elbows .... 354 

To construct a Pattern for a Right- 
angled Cylindrical Elbow .... 356 

To construct Oblique-angled El- 
bows .358 

To construct Right-angled Elliptic 
Elbows 359 

To construct Oblique-angled Ellip- 
tic Elbows 359 

To construct Right Semi-hyperbo- 
las by intersecting lines . . 355, 359 

To construct the Quadrant of a Cir- 
cle by intersecting lines .... 360 

To construct the Quadrant of a 
given Ellipse by intersecting 
lines 360 

To construct the Quadrant of a Cy- 
cloidal Ellipse by intersecting' 
lines 360 

To describe an Ellipse of given di- 
mensions by means of two Posts,, 
a Pencil, and a String 360 

To find the length of the circum- 
ference of a given Ellipse . . 207, 361 

To construct a Semi-parabola by 
intersecting lines 361 



CONTENTS. 



PAGE 

Ovals, to describe . 178, 350, 352, 353 

Of Circular Elbows 361 

Table applicable to Circular El- . 

bows 362 

To construct a Right-angled Circu- 
lar Elbow of 3, 4, 5, 6, 7, or 8 

pieces, &c 361 

To construct a Collar for a Cylin- 
drical Pipe of the same diameter 

as the receiving pipe 365 

To construct a Cylindrical Collar 
of a given Diameter to fit a Re- 
ceiving-pipe of a greater given 

Diameter 366 

To construct a Cylindrical Collar 
to fit an Elliptic-cylinder at ei- 
ther right section of the El- ' 

lipse 367 

To construct a Cylindrical Collar 
of a given Diameter, to fit a Cyl- 
inder of the same Diameter, at 
any given Angle to the side of 

the Cylinder 367 

To construct a Cylindrical Collar, 
or Spout, of a given Diameter, 
to fit a Cylinder of a greater giv- 
en Diameter, at a given Angle 
to the side of the Cylinder ... 368 
Of Spouts for Vessels .... 369 
Of Pitched or Bevelled Covers . . 370 
To construct a Bevelled Circular 
Cover of a given Rise and giv- 
en Diameter 371 

General Applications of Principles 
in Dynamics 327 

Heat, Sensible, Latent, Specific, 
&c 329 



PAGE 

To construct a Pattern for a Bev- 
elled Elliptical Cover of a given 
Rise to fit an Elliptic Boiler of 
given Diameters 371 

To construct a Bevelled Cover of a 
given Rise, to fit a False-Oval 
Boiler of given length and width 371 

Of Can-tops 372 

To construct a Can-top of a given 
Depth and given Diameters . . 372 

To construct a Can-top of a given 
Pitch, and given Diameters . . 373 

Of Lips for Measures .... 374 

To construct a Lip for a Measure, 
the Diameter of the Top of the 
Measure being given 375 

Of Sheet Pans . 375 

To cut the Corners for a Perpen- 
dicular-sided Sheet Pan ... 376 

To cut the Corners for an Oblique- 
sided Sheet Pan 376 

To construct a Heart, or Heart- 
shaped Cake-Cutter 376 

To construct a Mouth-piece for a 
Speaking-Tube 376 

To construct a Pattern for the 
Body of a Circular- bottomed 
Flaring Coal-Hod, all the curves 
to be arcs of circles 377 

Solders, Alloys, and Compo- 
sitions 379 



SECTION B. 

Reductions, Exchange, Invest- 
ments, Mixed Negotiations, 
etc., etc , , .... 61 



DEFINITIONS 

OF THE SIGNS USED IN THE FOLLOWING WORK. 



= Equal to. The sign of equality ; as 16 oz. = 1 lb. 

.-j- Plus, or More. The sign of addition ; as 8 -j- 12 = 20. 

— Minus, or Less. The sign of subtraction ; as 12 — 8 = 4. 

X Multiplied by. The sign of multiplication ; as 12 X 8 = 96. 

-f- Divided by. The sign of division ; as 12 -— 4 = 3. 

*r Difference between the given numbers or quantities; thus, 12 v^ 8, or 
8 ^r 12, shows that the less number is to be subtracted from the 
greater, and the difference, or remainder, only, is to be used ; so, 
too, height j- breadth, shows that the difference between the height 
and breadth is to be taken. 

1 II I Proportion ; as 2 14 11 3 16; that is, as 2 is to 4, so is 3 to 6. 
/s/ Sign of the square root ; prefixed to any number indicates that the 

square root of that number is to be taken, or employed ; as 
V64 == 8. 

^/ Sign of the cube root ; and indicates that the cube root of the num- 
ber to which it is prefixed is to be employed, instead of the num- 
ber itself; as y^/64 = 4. 

8 To be squared, or the square of; shows that the square of the number 
to which it is affixed is the quantity to be employed : as 12 2 -f- 
6 = 24 ; that is, that the square of 12, or 144 ~ 6 = 24. 

3 Indicates that the cube of the number to which it is subjoined is to 
to be used ; as 4 3 = 64. 

« Decimal point, or separatrix. See Decimal Fractions. 

Vinculum. Signifies that the two or more quantities over 

which it is drawn, are to be taken collectively, or as forming 
one quantity ; thus, 4 -{- 6 X 4 = 40 ; whereas, without the 
vin culum, 4 + 6 X 4 = 28 ; also, 12 — 2X3+4 = 2 ; and 
V"52^T32 ^ 4 go, also, V (5 2 — 3 2 ) = 4, and (4 -f- 6) X 4 
= 40. 

4 2 ( half of 4 2 or ) 

2 \ half of the square of 4 ) ~~ 

(A2 \2 
— J (the square of half the square of 4) = 64. 

4& 2 or £ {b) 2 (half the square of b.) 
{hby (the square of half b > 
(26) 2 (the square of twice b.) 



SECTION A. 

FOREIGN MONEYS OF ACCOUNT, COINS, WEIGHTS, 
AND MEASURES, 

REDUCED TO THEIR VALUES IN THE MONEY, WEIGHTS, AND 
MEASURES OF THE UNITED STATES. 

The many changes that have been made in the moneys of ac- 
count, coins, weights, and measures of different countries, by 
their respective governments, within the last few years, chiefly, 
though not in all cases, by the adoption of the Metric System, or 
systems bearing aliquot relations thereto, have compelled the au- 
thor to re-write this section of the work, in a great measure, since 
the first edition was published ; and it is the intention that this 
edition, and subsequent editions that may be published, shall con- 
tain this section strictly correct in all particulars at the time of 
going to the press. 

The Federal units of comparison in the following tables, unless 
otherwise expressed, are as follows; viz., the dollar of 100 cents, 
in gold ; the commercial or avoirdupois pound, of 7000 grains ; the 
commercial yard, of 36 inches ; the commercial or wine gallon, 
of 231 cubic inches; the commercial or Winchester bushel, of 
2150 t 4 q 2 q cubic inches; the standard foot, of 12 inches; the statute 
mile, statute acre, &c. 

The value in Federal money, therefore, affixed to any particular 
denomination of a foreign money of account in the following 
tables, is the equivalent, or intrinsic par, of that denomination 
in United-States gold coins. It is predicated upon the standard 
weight and purity of the coins coined especially to represent that 
denomination, or conventionally held to be the measure of its 
value, compared with the standard weight and purity of the gold 
coins of the United States, that represent the dollar or its multi- 
ples. 

Thus, in respect to those countries in which gold is made the 
measure of value and chief legal tender, it is the intrinsic par, 
, gold for gold ; and, in respect to those countries in which silver is 
1 l 



AOL FOREIGN MONEYS OF ACCOUNT. 

made the measure of value and chief legal tender, it is the par 
value of that denomination in United-States gold coins, based 
upon the almost constantly prevailing relative commercial values 
for many years past, of gold to silver, as 15§ to 1, for equal weights. 
It is, therefore, in a commercial point of view, the intrinsic par of 
that denomination, in Federal gold coins, in all cases. 

The denomination itself, to which the Federal value is imme- 
diately affixed, is usually the integer, or ultimate money of ac- 
count, of the country especially referred to. It is a money of 
account in that country always, but not always the name of a cir- 
culating coin. Occasionally, even, its value is not represented by 
any known single circulating coin. 

From the foregoing remarks, it will be perceived that, when the 
mintage relative values of gold to silver, in any particular country, 
are maintained at rates nearer to each other than 15-| to 1 for 
equal weights, the gold coins of that country are commonly worth 
more, as commercial material, than its silver coins of the same .de- 
nominations, or same prescribed values ; and, conversely, that when 
the mintage relative values are limited to rates more remote from 
each other than 15§ to 1, the gold coins are commonly worth less 
than the silver coins. 

Thus, the Federal dollar, in standard silver coins, is ordinarily 
worth, as commercial material, but 14.88372 -f- 15.375' — 96| cents 
in Federal gold coins. But, since most Governments make silver 
the chief measure of value, the mint value of gold is usually pur- 
posely placed above its commercial worth. Thus, twenty francs 
in French gold coins are ordinarily worth, as commercial material, 
but 15.375 X 20 4- 15.5 = 19.8387 francs in French silver coins. 

It is true that the Silver coins of the United States, in small 
sums, for immediate use, in limited localities at home, may occa- 
sionally sell in exchange for the gold coins at their nominal values, 
or even at a premium, according to the local demand and supply ; • 
and the same may happen with regard to the gold coins in ex- 
change for the silver, in France, and those other countries where 
gold is purposely over-valued in the mintage ; but these conditions 
do not affect the general commercial relations of the metals : they 
are due only to a slight derangement in the required distributions 
of the two kinds of coins. 

In Germany and Austria, the mint relative values of gold and 
silver for the Zollverein money, are as 15f to 1, for equal 
weights. 



FOREIGN MONEYS OF ACCOUNT. a 3 



FOKEIGN MONEYS OF ACCOUNT AND COINS KEDUCED TO 
THEIR VALUES IN FEDERAL MONEY. 

Foreign. IT. States. 

ABYSSINIA, (E. Africa). — Massuah: The old 
Venetian zechino (sequin) is current here at 50 
harfs ; and 23 harfs = 1 pataka, or old Spanish 
dollar, - - - - - = $1.01385 

Austrian rix-dollars and Spanish dollars are cur- 
rent here at 1 pataka each. 

Note. — The old peso duro colonato, or Carolus silver dollar of Spain, con- 
tained, at mint usage, 415 grains of mint silver -|~| fine = $1.041353; but it is no 
longer struck at the mint, and those in circulation are more or less abraded. It 
is now valued, throughout the British Possessions in North America, and gen- 
erally, wherever it circulates by tale, or is made the integer of the moneys of 
account, at 50 pence sterling in gold = $1.0138542. The Austrian rix-dollar 
(tallaro), scudo, or crown, which, by the way, has not been coined since 1858, 
except on orders for foreign circulation, contains yL- Vienna mark of fine silver, 
or 361.11 grains = $1.0114911. This is often called the German dollar; and the 
Venetian dollar is of the same value. 

ALGERIA (N. Africa). — A Igiers, Bona, §•<?.: 100 

centimes = 1 Franc - =0.19452 

ARABIA. — Muscat: 20 goz = 1 niamooda, 20 m. = 

1 current Spanish dollar . - - - =1.01385 

Note. — 1 goz = 2 paras, and 1 mamooda = 1 piastre of Egypt. See Egypt. 

Mocha, Hodeida: 2 crats = 1 commasse; 60 c. = 1 
Mexican or Spanish dollar by tale = 360 grains 
of fine silver," - - - - - = 1.00838 

160 crats = 1 wakega or troy ounce (gold and 
silver weighty. 
'Jidda: Same as at Alexandria, Egypt. 
Aden : 80 caveers ±± 1 piastre of account = -| cur- 
rent Spanish dollar,* - - - - = 0.84488 
Also, as at Calcutta. Official, as in Great Britain. 
AUSTRALIA. — Sidney, Melbourne, Hobart Town, 
and Australasia generally : 
Standard of purity, denominations, values, and 
relative values, since 1855, same as in Great 
Britain. 
AUSTRIA. — Vienna, Prague, Trieste, Ragusa, fyc. 
Zollverein money: Standard for gold and sil- 
ver coins = T 9 ^- fine, each; relative values, 
gold to silver as 15.375 to 1. 



4« FOREIGN MONEYS OF ACCOUNT. 

Foreign. U. States. 

4 pfenninge == 1 kreuzer ; 60 k. = 1 gulden, or 
florin = -^g- Zollverein pfund (11^ grams) of 
fine silver, or 171.471 grains, - - - = $0.4803 

\% gulden = 1 Zollverein thaler; 1107 gulden 
= 80 Zoll. krones ; 81 gulden = 200 francs ; U. 
S. Customs value of gulden = 
AZORE ISLANDS. — Fayal, Terceira, Corvo, St. Mi- 
chael, fyc. : 

1000 reis = 1 milreis of account = -| old cur- 
rent Spanish dollar, - - - - = 0.84488 

U. S. Customs value = 83^- cents. 

Note. — In 1834, English sovereigns and Spanish dollars were made legal 
tender here and at the Madeiras ; the former at the rate of 4120 reis, and the lat- 
ter at 870 reis, each, which corresponds very nearly with their intrinsic values 
in standard Portuguese gold coins. 

BALEARIC ISLANDS. — Majorca, Raima; Mi- 
norca, Port Mahon : Same as new system in 
Spain, see Spain. 

BELGIUM. — Brussels, Antwerp, Ostend, fyc. . Stan- 
dard for gold and silver coins = -^ fine, each. 
Relative values, gold to silver as 15.8228* to 1. 
100 centimes = 1 Franc = 4^ grammes of fine 

silver, ______ —0.19452 

BERBER A (E. Africa.) : Same as at Mocha, 
Arabia. 

BERMUDA ISLANDS.— Official, as in Great Brit- 
ain. In trade, 100 cents = 1 dollar = 1 old 
current Spanish peso, or 50 pence sterling in 
gold, - - - - - - =1.01385 

Note. — At the Bermudas, in British America, and other British foreign pos- 
sessions generally, official or government accounts are kept, and duties to the 
government are assessed, in sterling money; hut until 1842 this class of accounts 
were kept at the Bermudas and Jamaica, in pounds, shillings, and pence, at 12 
shillings sterling to the pound, when it was ordered that hereafter they be kept 
in sterling money, and that all existing contracts in those colonies be settled at 
the rate of 4 pound sterling per colonial pound. 

BOURBON ISLAND. — Si. Denis : 100 centimes = 1 

franc - . - - - - - =0.19452 

* Although the silver coins of Belgium, both by law and general usage, have 
the same intrinsic values as those of France of the same denominations, yet this 
rule does not hold good with regard to the gold coins. The mint standard for 
25 francs of France is 8i&- grammes of mint gold -JKr fine, while that for 25 
francs of Belgium is only 7 -A- grammes of mint gold -^ fine ; and so in propor- 
tion fqr the other gold coins. 25 francs, French mint, are worth $4.823816, while 
the 25-franc piece, Belgic mint, is worth only $4.72541; in other words, the Bel- 
gic gold coins are less in value two centimes per franc than the French gold 
coins. 



FOREIGN MONEYS OF ACCOUNT. «5 

Foreign. U. States. 

CANADA, DOMINION OF, and British America 
generally : Standard for silver coins (20-cent 
pieces or Colonial shillings) = f | fine ; for gold 
coins (British sovereigns) = \\ fine. Relative 
values, gold to silver as 14.341 to 1. 
100 cents = 1 dollar colonial, - =$1.00 

Also, 4 farthings = 1 penny ; 1 2 p. = 1 shilling ; 

20 s. = 1 pound colonial, - - - = 4.00 

Note. — The standard 20-cent piece, or colonial silver shilling, of British 
America, contains 66|- grain's of fine silver, and is, therefore, worth only 18.655 
Federal cents in Federal gold coins, or 19.2708 Federal cents in Federal silver 
coins. But the money of account shilling in British America is equal to 20 Fed- 
eral cents in gold; thus, the British pound sterling in gold, the British sovereign, 
is equal to $4.8665 in gold; and the shillings in that sovereign are equal to 
24.3325 Federal cents, each, in gold; therefore, ^^ffMlHP — $ 4 - 0() > the 
value of the colonial pound in Federal money (gold), measured hy, or payable 
in, British standard gold. 

CANARY ISLANDS. — Teneriffe, Palma, Grand 
Canary, Fuerteventura, &fc. : Official as in Spain ; 
in trade, occasionally, 8 reals (antiquas) of 34 
maravedes each = 1 piastre, or peso of ex- 
change, __-'._ —0.75623 

CANDI ISLAND. — Same as in Turkey. 

CAPE OF GOOD HOPE (S. Africa). — Cape 
Town, 8fc. : .Same as in Great Britain. 

CAPE VERDE ISLANDS. — St. Vincent, Min- 
dello ; St. Jago, Porto Praya, &fc. : 
1000 reis = 1 milreis. Old Spanish dollars are 
current here at 870 reis. 

Central and South America. 

CENTRAL AMERICA. — Honduras, Truxillo, 
Port San Lorenzo, Omoa, 8fc; Nicaragua, 
Realejo, Greytown, Sec; San Salvador, La 
Union, Sonsonate, Sfc; Costa Rica, Puntas 
Arenas, Matina, 8fc. : Guatemala, Ystapa, fyc. : 
Standard for silver coins (dollars) =z^j Castil- 
ian marco of silver -|-^i fine, or -^ marco of fine 
silver to the dollar ; for gold coins (double escu- 
dos of- 32 reals) = y-fy Castilian marco of gold 
l^i fine. Relative values, gold to silver as 
16.5614 to 1. 

100 centavos or 8 reals = 1 dollar = 355.08 

grains of fine silver, - - - - = 0.9946 

Spanish dollars and U. S. gold coins circulate 
here, dollar for dollar. 
1 * 



6a 



FOREIGN MONEYS OF ACCOUNT. 



K97461 



= 1.01385 



Foreign. U. States. 

Balize, Balize: Official accounts are kept here in 

sterling money. 
SOUTH AMERIC A. — Peru, Callao, May, Trux- 
itto, Arica, Sfc; Chili, Valparaiso, Conception, 
Coquimbo, Sfc. ; New Granada, Cartagena, 
Santa Martha, Savanillo, Buenaventura, Sfc.; 
Ecuador, Guayaquil, fyc. The prescribed 
standard for the mintage of these States is now 
in conformity with that of France, and there is 
strong probability that Brazil, and the other 
States in South America having mints, will 
soon adopt the same standard. Standard for 
gold and silver coins = T 9 g- fine, each. Relative 
values, gold to silver as 15-1- to 1. 

100 cents == 1 sol, or dollar = 22i grammes of 
fine silver, - - - - -: 

Also, 100 centesimas = 1 duro or old Spanish 
dollar, ----- 
Brazil. — Rio Janeiro, Maranham, Baliia, Para, 
Pernambuco, $*c. : Standard for silver coins = 
y 1 ^ Castilian marco of silver l^ fine per milreis ; 
for gold coins = T |-g Castilian marco of gold 
-^ fine per 10 milreis. Relative values, gold to 
silver as 14.30556 to 1. 

1000 reis = 1 milreis = 180.8278 grains fine sil- 
ver, _--.-. —0.50651 

A current Spanish dollar passes for 2 milreis of 
account, and the modern gold coins (10 milreis 
and 20 milreis) = $0.544375 per milreis. 
Bolivia. — Cobija, Sfc. : Standard for silver coins 
(dollars) — -^ marco of silver A-| fine ; for gold 
coins (doubloons) = T 2 y marco of gold -| fine. 
But little gold is coined in Bolivia, and that in 
circulation has, at present, no nominal mint re- 
lation to the silver coins. 

100 centavos = 1 dollar = 283.03478 grains of 

fine silver, - =0.79292 

Note. — The gold coins of Bolivia are often light of weight, and seldom 



The fractional silver coins are worth 



are often hut little if any ahove -^ fine, 
less, relatively, than the integer. 

Venezuela. — La Guayra, Maracaybo, Cumana, Pu- 
erto Cabello, Syc. : In Venezuela, as in Peru, &c, 



FOREIGN MONEYS OF AC.COUNT. «7 

Foreign. U. States. 

the silver 5-franc piece of France is made the 
measuring unit of value, and is divided into 100 
centavos. In the moneys of account, however, 
the peso macuquins of 80 centavos is sometimes 
used ; and one dollar in United-States gold coin 
is assumed to be worth 7 cents more than 5 sil- 
ver fsancs, which is the case at the metal ratio 
of gold to silver as 16.0794 to 1, for equal 
weights. Hence, 1 peso fuerte Americano de 
premio = \ / = $1.3375, measured by the 
peso macuquins. 
Argentine Republic. — Buenos Ayres, Parana, Sfc; 
Uruguay, Montevideo, fyc. ; Paraguay, As- 
sumption, Neembucu, Sfc. : 

Foreign gold and silver coins circulate here meas- 
ured generally by the Spanish dollar : 10 deci- 
mos = 1 real ; 8 r. = 1 peso. 
Guiana, Cayenne : Same as in France. 

Paramaribo : Same as in Holland. 

Georgetown : Same as in Great Britain. 
Falkland Islands. — Same as in Great Britain. 
CHINA. — Canton, Shanghai, Amoy, Ningpo, Foochoo, 
Hongkong L, Macao : 

Standard for gold and silver ingots = 94 touch, or 
■^q fine, each. . Gold is not treated as money, 
but as merchandise. 

10 cash or le = 1 candarine or fun ; 10 c. = 1 
mace or tseen; 10 m. = 1 tael or leang == 
583^- grains of silver T 9 ^ fine, or 548^- grains of 
fine silver, - -=$1.53591 

Note. — Mexican dollars, which are about the only coins, except copper cash, 
that circulate in China, on account of their convenience generally, bear a pre- 
mium well laid on, upon their intrinsic worth. They commonly pass for about 
72 candarines each, or j-£q of the tael in ingots of standard purity. A tael 
weight, or leang as it is called by the natives, of wan-yin or sysee (see-sze) 
silver, that is, of silver as fine as silk, or that may be drawn into a thread as 
fine as silk (pure silver), is valued by the East India Company at 80 pence 
sterling. In London, it is commonly valued by the price per ounce paid at the 
mint for the fine silver in foreign silver coins, which is about 66 pence sterling, or 
" SJ "fw^ — 8®-% P ence sterling per tael. In 1799, the Congress of the United 
States fixed the value of the tael, in ingots of standard purity, for customs' pur- 
poses, at $1.48; but this was when silver was the measure of value in the 
United States, and when the dollar contained 371| grains of fine metal. The 
law, it is believed, has not been repealed. 

CORSICA ISLAND. — Ajaccio : 100 centimes = 1 

franc, ------ =0.19452 



8 a FOREIGN MONEYS OP ACCOUNT. 

Foreign. U. States. 

CYPRUS ISLAND. — Same as in Turkey. 
DENMARK. — Copenhagen, Elsinore, Odense, fyc. : 
Standard for silver coins — j^-q fine. 
1 6 skillinge = 1 mark ; 6 m. 1 rigsdaler = y|f ■§■ 
Cologne mark of mint silver = ^ former spe- 
ciesdaler = 195.1233 grains fine silver, - = $0.546552 
U. S. Customs value of speciesdaler = $1.05. 

Note. — Denmark coins ducats and 5-thaler and 10-thaler gold pieces; but 
these and the speciesdaler have no proposed mint-relation to each other. The 
ducat is of the ordinary value of that denomination of coins ; and the 10-thaler 
piece is designedly and practically of the same intrinsic value as that of North 
Germany; viz., equal to ten Bremen thalers in silver at the metal ratio of 15.409 
to 1; or equal, at the common par of exchange, to $7.90005 in gold. 

EGYPT (N. Africa). — Cairo, Alexandria, Suez, 
fyc.: 40 paras = 1 piastre ; in current gold coins 
= $0.04996647 ; in current silver coins = 
$0.05054979. 

Note. — In 1836, British sovereigns were made legal tender throughout Egypt 
at 97 pi., 20 pa., each ; Spanish doubloons or onzas at 313 pi., 29 pa., ; Napoleons 
(20 francs in gold) at 77 pi., 6 pa. ; Venetian sequins at 48 pi., 13 pa. ; Dutch du- 
cats at 45 pi., 26 pa. ; tallaros (G-erman dollars) at 20 pi. ; colonatos (Spanish dol- 
lars) at 20 pi., 28 pa.; 5-franc pieces at 19 pi., 10 pa. The measure of value is 
|- oka-drachmas of fine silver to the piastre = 2-§- Abyssinian dirhem or 18 troy 
grains = $0.050419109. 

FRANCE. — Standard for gold and silver coins — t 9 q 
fine, each. Relative values, gold to silver as 
151 to 1. 

10 centimes = 1 decim; 10 d. = 1 franc = 4-| 

grammes of fine silver, - - - = 0.19452 

In practice, 100 centimes = 1 franc. U. S. Cus- 
toms value = $0,186. 
GERMANY (The Zollyerein States, or North Ger- 
man Confederation) : Prussia, Saxony, 
Mecklenburg, Oldenburg, the Hanse 
Cities, 8fc. Zollverein money: Standard for 
gold and silver coins = T 9 ^ fine, each. Rela- 
tive values, gold to silver as 15.375 to 1. 

12 pfennige = 1 silber groschen ; 30 s. g. = 1 - 
thaler = -^ Zollverein pfund (16§ grams) of 
fine silver, or 257.2058 grains, - -=0.72045 

369 Zollverein thalers = 40 Zollverein krones; 
1 Zoll. krone = -^ Zollverein pfund (10 grams) 
of fine gold, or 154.3235 grains == $6.64614 ; 
27 thalers = 100 francs. 

U. S. Customs value of thaler = 
Note. — Saxony reckons 10 pfennige to the groschen. 



FOREIGN MONEYS OF ACCOUNT. a 9 

Foreign. U. States. 

Bremen (special): 5 schwaren = 1 groot; 72 g. = 

1 thaler = \ old Frederic d'or, - -=$0.79619 

U. S. Customs value = 78| cents. 
Hamburg, Lubec, Altona (special) : 12 pfennige = 
1 schilling; 16 s. = 1 mark. 
1 marc current = -^ Cologne mark of fine silver, _= 0.28869 
1 marc banco, at par (London rate of 13m. 12s. 

to the £.), - - - - = 0.35393 

U. S. Customs value of mark current = 28 cents ; . 
of mark banco = 35 cents. 
Southern States. — Bavaria, Wurtembtjrg, Ba- 
den, Sfc. : Standard for silver coins = -^ fine. 
4 pfennige = 1 kreuzer ; 60 k. _z_ 1 gulden or 
florin == -jf-g- Zoll. pfund of fine silver (9-|l 
grams) or 146.975 grains, - - - =0.41169 

7 gulden = 4 Zollverein thalers; 189 gulden = 

400 francs. 

U. S. Customs value of gulden = 40 cents. 

GREAT BRITAIN. — Sterling money : Standard for 

silver coins = -f % fine ; for gold coins = \\ fine. 

Relative values, gold to silver as 14.287891 to 1. 

4 farthings = 1 penny; 12 p. 1 shilling ; 20 s. = 

1 pound, _____ =4.86656 

U. S. Customs value = $4.84. 
Note. — Since 1816, the Government of Great Britain has estimated the value 
of gold compared with that of silver as 14-^-|-||j- to 1, for equal weights. In 
theory, by the mint regulations, &pound sterling is equal to 1614y 6 Y grains of 
fine silver, or HS-g-^g- grains of fine gold. A pound sterling in silver, there- 
fore, measured by the Federal standard of 345-| grains of fine silver to the dol- 
lar, is equal to $4i-§-|- in Federal silver coins; or a standard silver shilling, 
measured by this measure, is equal to 23^ 7 -<4g cents; and at the ordinary par of 
exchange it is worth $0.226122, very nearly, in Federal gold coin. But a pound 
sterling in British gold (a sovereign) measured by the Federal standard of 2344- 
grains of fine gold to the dollar, is equal to $4ff§|§|, or $4.866563529, 
very nearly, in Federal gold coins. At the former relative values of the two 
currencies, or relative values before the Federal standard was changed, in 1834, 
viz., 4 shillings and 6 pence sterling to the dollar, or $4.44|- to the pound, the 
par of exchange is .09498, or 9-| per cent, practically, in favor of sterling money. 
Gold is the measure of value in Great Britain, as well as in the United States; 
and the silver coins of that government are not legal tender at home in sums ex- 
ceeding £2. 

GREECE. — Athens, Patras, the Ionian Islands, Sfc: 

Standard for gold and silver coins = T 9 7 fine, 

each. Relative values, gdld to silver as 15.549 

to 1. 

100 lepta = 1 dramia = 1 dramia weight of fine 

silver, or 621 grains, _ = 0.17401 



10 a FOREIGN MONEYS OP ACCOUNT. 

Foreign. U. States. 

5 dramias = 1 taliron. 20 dramias (gold) = 1 
othonion = $3.44137. 

100 oboli =1 Spanish, German, or Venetian dol- 
lar = $1.01385. 

Note. — G-reece is a party to the present movement (1869) to legalize and en- 
force the use of the metric system of moneys, weights, and measures, through- 
out Continental Europe. ' 

HOLLAND (Netherlands). — Amsterdam, Rotter- 
dam, The Hague, Sfc. : 

Standard for silver coins == ^f fine ; for gold 
coins — T 9 7 fine. Relative values, gold to silver 
as 15.7333 to 1. 

100 centimes = 1 guilder or florin = 10 grams of 

mint silver, - -=$0.40806 

Gouden Willem (10 guilders) = 6§ grams of mint 
gold .= $3.98769. 

U. S. Customs value of guilder = 40 cents. 
HAWAIIAN or SANDWICH ISLANDS. — Hono- 
lulu, Sfc. : 

100 cents = 1 dollar, - - - - = 1.00 

India and Malaysia, or East Indies. 

Hindostan. — Bengal, Madras, Bombay, Presi- 
dencies of: Standard for gold and silver coins 
= \\ fine, each. Relative values, gold to silver 
as 15 to 1. 
Calcutta, Madras, Rangoon, fyc* : 1 2 pice = 1 anna ; 
16 a. = 1 rupee =1 tola, or 180 grains of mint 
silver, = \^ old sicca rupee, - - - •= 0.46217 

1 mohur, or gold rupee (E. I. Co.), = 15 silver 
rupees in theory = 1 tola of mint gold, or 165 
grains of fine gold, = $7.10594. 
Bombay, Surat, Sfc. : 100 reas = 1 quarter; 4 q. = 

1 rupee, - - - - - =0.46217 

U. S. Customs value of rupee = 441 cents. 
Madras, fyc. (old usage) : 48 jittas = l fanam; 36 f. 
= 1 star pagoda = 4 arcot, or Company ru- 
pees, - - - - - - = 1.84869 

48 jittas = 1 fanam ; 36 f. = 1 India pagoda = 8 

shillings sterling in gold, - - - = 1.9466 

U. S. Customs value of star pagoda = $1.84; of 
Indian pagoda, = $1.94. 
Goa : Official, same as in Portugal. 
Pondicherry : Same as in France. 



FOREIGN MONEYS OP ACCOUNT. 



all 



Foreign. U. States. 

Ceylon Island. — Colombo, Trincomalee, Sfc. : Offi- 
cial, same as in Great Britain ; also, 1 rixdaal- 
der= 18c?.; 1 Spanish dollar = 50c?. ; 1 rupee 
= 22c?., sterling. 
Malaya. — Malacca, Pahang, Perak, fyc. : Same as 

at Singapore. 
Penang Island. — Same as in Calcutta; also, 100 

cents == 1 Spanish dollar. 
Siam. — Bangkok, 8fc. : 4 prangs, or clams, = 1 som- 
pay ; 4 s. = 1 salung ; 4 salungs = 1 tical = 
1 tical weight of silver 93 touch, or ^\ fine, -=$0.61659 
U. S. Customs value of tical = 61 cents. 
Singapore Island. — Singapore: 100 cents = 1 

dollar (old Spanish), - - - - = 1.01385 

Banca Island. — Same as at Batavia (Java I.). 
Borneo Island. — Banjermassin, Sarawak, fyc. : 

Generally as at Batavia. 
Celebes Island. — Macassar, and the other Dutch . 

settlements : Same as at Batavia. 
Java Island. — Batavia, Samarang, fyc: 100 cents 

= 1 guilder, - - - - - = 0.40806 

Mexican dollars are current here at 2f guilders, 
each. 
Molucca Islands. — Amboyna, §•<?.: Same as at 

Batavia (Java I.). 
Philippine Islands. — Luzon, Manila, Mindano, 
fyc. : Official, as in Spain ; also, 34 maravedis 
== 1 real ; 8 r. = 1 peso, or dollar, - -== 1.00465 

Sumatra Island. — Bencoolen, Padang : Official, as 
at Batavia ; also, 8 satellers = 1 soocoo ; 4 s. 
= 1 dollar, or rial, = 1 pardow of Acheen. 
Acheen : 16 copangs= 1 mace ; 10 mace of fine sil- 
ver, or 374.306 troy grains = 1 pardow, -== 1.04845 
Mexican and Spanish dollars pass for 1 pardow, 
each. 
ITALY. — The recent formation of the several States in Italy in- 
to one kingdom, Rome and in its immediate vicinity only excepted, 
has had the effect to unify the moneys and moneys of account in 
that portion of Europe also ; and now, throughout "Italy proper, 
or the Kingdom of Italy, including the Island of Sicily and that 
of Sardinia, the official monetary system, with a slight difference 
in nomenclature, is identically the same as that of France. 
Standard for gold and silver coins = -^ fine, each. 
Relative values, gold to silver as 15^ to 1. 



12« FOREIGN MONEYS OF ACCOUNT. 

Foreign. U. States. 

100 centesimi = 1 lira, or franco, = 4-| grams of 

fine silver, - =-$0.19452 

JAPAN. — Yeddo, Miaco, Osaka, Simoda, Halcodadi, 
Nagasaki, Yokohama, Matsmay, Napa, fyc. : 
New System : Standard for silver coins (assay) 
= T 8 ¥ 9 Q fine ; for gold coins = -|f f fine. Rela- 
tive values, gold to silver (assuming the cobang 
to represent 10 silver itzebous) as 15.01877 to 1. 

50 sen = 1 itzebou, or itchibou (often called boo} 
= 5 monme of mint silver, or -|f monme of fine 
silver = 119.1888 grains, - - - =0.333855 

1 cobang contains |l| monme of mint gold, or 

79.35992 grains of fine gold, - =3.417744 

U. S. Customs value of itzebou = 

Note. — Mexican dollars circulate in Japan, commonly at 2.87 to 2.90 itzebou, 
each; but they are intrinsically worth over 3 itzebous, each, by tale. 

LIBERIA (W. Africa). — Monrovia, 8fc: Same as 

in the United States. 
MADEIRA ISLANDS. — Funchal, frc: 1000 reis 

= 1 milreis of account, - - -=1.01385 

See Azore Islands, note relative to. 
U. S. Customs value of milreis = Si. 00. 
MALTA ISLAND. — Valetta, frc. : Official, as in 
Great Britain; also, 20 grani = 1 taro ; 12 t. 
= 1 scudo = l ducat of Naples, - - = 0.41371 

U. S. Customs value of scudo = 40 cents. 
MAURITIUS ISLAND.— Port Louis, &c: Official, 
as in Great Britain ; also, 100 .cents = 1 dollar. 
Tallaros of Austria and silver Napoleons (5- 
franc pieces of France) are current here at 1 
dollar each; Spanish pesos and Mexican dol- 
lars, 'at 52 pence sterling each. 
MEXICO. — Acapulco, Mazatlan, San Bias, Cam- 
peachy, Tampico, Sisal, Vera Cruz, fyc. :■ 
Standard for silver coins (dollars, average by as- 
say) — 416.15 grains -^ fine; for gold coins 
(doubloons, average by assay) =416.4 grains 
fff fine. Relative values, gold to silver as 
16.618192 to 1. 
100 cents = 1 dollar; also, 6 grani = 1 cuarto; 
2 c. = 1 medio ; 2 m. = 1 real ; 8 r. = 1 dollar 
= 374.535 grains of fine silver, - - = 1.0491 

Note. — The Mexican marco = 3549.81 grains; and the standard silver dollar 
should weigh ma^jsASL = 416.4 grains. 



FOREIGN MONEYS OF ACCOUNT. a 13 

Foreign. U. States. 

MOROCCO (N. Africa).— Morocco, Fez, Tangier, 
fyc. : 24 fluce = 1 blankeel ; 10 b. = 1 metical ; 
4m,z:l oncia or ducat = 4 meticals weight of 
fine silver, or 295.3846 grains = 1 old standard 
ducat of Naples, - - - -=$0.82739 

MOZAMBIQUE (E. Africa). —Mozambique, Quit- 
imane, Sofala, Delagoa Bay, fyc. : 1000 reis = 1 
milreis of account. 

Note. — A variety of foreign coins are current here, and many of them wide 
of their true values; viz., Spanish and Mexican dollars at 1000 reis, each; silver 
5-franc pieces of France and United-States gold coins per dollar at 900 reis, each; 
Spanish doubloons at 17120 reis, patriot doubloons at 16000 reis; and British 
sovereigns at 4500 reis, each. 

NORWAY. — Christiania, Bergen, 8fc. : Standard for 
silver coins = -^^ fine. 
20 skillinge == 1 mark ; 6 m. = 1 speciesdaler = 
2 rixdalers = ^ 5 -g Cologne mark of mint silver 
or 390.2465 grains of fine silver, - - = 1.0931 

Note. — Norway and Sweden are about to coin gold of 10 and 20 francs. 

NUBIA (E. Africa). — Suakin, Sfc: Same as at 
Alexandria," Egypt. 

PERSIA. — Busliire, Gombroon, Astrabad, fyc: 100 
mamoodis, or 50 abasse, = 1 toman = f miscal 
weight of fine gold, or 49.23077 grains, - = 2.12019 

5 dinars = 1 kasbequis ; 2 k. = 1 dinars-biste ; 2 
d.-b. = 1 shatree, shafree, or shahis ; 2s. = 1 
mamoodi ; 2 m. = 1 abasse ; 2% a. = 1 papa- 
bat ; 2 p. = 1 saheb-keran. 

Note. — Foreign gold and silver coins in great variety circulate in Persia, but 
at fluctuating prices; the toman, however, is commonly valred at two Mexican 
dollars. 

PORTUGAL. — Lisbon, Oporto, St. Ubes, §•<?.; Stan- 
dard for gold coins (coroas) == -^ marco of 
gold \\ fine ; for silver coins (crusados) = ^ 
marco of silver ^ fine. Relative values, gold 
to silver as 14.72524 to 1. 

1000 reis = 1 milreis = 2 crusados = 398.423,7 

grains of fine silver, - -==1.11601 

1 milreis in gold = $1.16525. 

Method of writing and reading monetary quanti- 
ties: Example. — Rs. 5 : 600© 750 = 5,600: mil- 
reis and 750 reis. 

U. S. Customs value of milreis ==$1JL 2. 

Note. — In commercial transactions, lately, Mexican dollars commonly pass 
for 1 milreis each, and United-States gold coins at 970 reis per dollar. 



14« FOREIGN MONEYS OF ACCOUNT. 

Foreign. U. States. 

ROME and Clvita Vecchia: Standard for gold and 
silver coins = T 9 F fine, each. Relative values, 
gold to silver as 2±q*±£» = 15.43367 to 1. 
5 quatrini = 1 baiocho; 10 b. = 1 paolo ; 10 p. 

= 1 scudo, or crown, = 24^- grams of fine silver =$1. 04609 
The gold coins are 1\, 5, and 10 scudo pieces, and 

contain i|| grams of fine gold per scudo. 
U. S. Customs value of scudo = $1.05. 
RUSSIA. — St. Petersburg, Riga, Cronstadt, Odessa, 
Sfc. : Standard for silver coins = -| fine ; for gold 
coins (assay) = |-||- fine. Relative values, gold 
to silver as 15.17295 to 1. 
10 kopecks = 1 grieven ; 10 g. = 1 rublyu (rou- 
ble) = 4 2 5°r funt of fine silver, or 280.19 grains = 0.78483 
In practice, 100 kopecks = 1 rouble. 
100 silver roubles = 360 bank or paper roubles. 
Bank rouble varies from par to 4 per cent pre- 
mium. 
U. S. Customs value of silver rouble = 75 cents. 
SENEGAMBIA (W. Africa) .— Gambia, Bathurst, 
Sierre Leone, Sfc. : Official, as in Great Britain. 
St. Louis : Official, as in France. 
SOCOTRA ISLAND. — Same as at Muscat, Arabia. 
SPAIN. — Madrid, Malaga, Cadiz, Santandre, Bilboa, 
Barcelona, Sfc. : New system, legalized Oct. 19, 
1868, and its use made obligatory to the exclu- ' 
sion of all other systems after Dec. 31, 1870. 
Standard for gold coins = ^ fine ; for silver 
coins (5-peseta pieces, duros, or pesos) = T 9 ^- 
fine ; 1-peseta pieces and less = -jjr§-£ fine. Rel- 
ative values, gold to silver (duros) as 15^ to 1. 
100 centesimas = 1 peseta = 41 gramos of fine 

silver = 1 franc of France, - - -=0.19452 

Note. — In this system the peseta of account is equal to 1 silver franc of 
France ; and the 5-peseta silver coin is equal to 5 francs in silver of France ; but 
the 1-peseta coin is worth hut $0.18047. The gold coins are worth peseta for 
franc in gold. 

Prevailing system last preceding the foregoing : Stan- 
dard for silver coins (escudos of 1 reals velldn) 
= T 9 7 fine; for gold coins (10 escudos and over) 
= ii~| fine. Relative values, gold to silver as 
15.5555 to 1. 
1000 milesimas = l escudo = -^ mareo of fine 

silver, or 17 9j grains, - -=0.50232 



FOREIGN MONEYS OF ACCOUNT. «15 

Note. — The silver coins of less denominations than 4 reals, and the gold 
coins of less denominations than 10 escudos, have less purity than those mentioned 
above; the 80-reals piece, even, is hut ^f-f- fine = $3.85446, while the 100-reals 
piece (double de Isabel) is worth $4.96493. 

Foreign. U. States. 

Gibraltar : 16 quartos = 1 real ; 12 r. = 1 peso duro 

or old Spanish dollar, - - - -=$1.01385 

SWEDEN. — Stockholm, Gothenburg, Carlscrona, Gefle, 
fyc. : Standard for silver coins = -| fine ; for 
gold coins (ducats) — J# fine. Eelative values, 
gold to silver as 15.18517 to 1. 
100 ore = 1 riksmynt, or riksdollar riksgald = i 

speciesdollar, - - - - - ■== 0.27622 

1 speciesdollar = -$fe mark of fine silver, or 

393.15 grains, - -=1.10124 

U. S. Customs value of speciesdollar = $1.06. 

1 ducat = 8 riksmynts, nominally, == ^f-^ marks 
of fine gold == $2.23738. 

Sweden is about to coin gold coins of 10 and 20 
francs. 
SWITZERLAND. — Basel, Bern, Geneva, Lausanne, 
Lucerne, Neufchatel, Zurich, fyc. : 
Standard for gold and silver coins = T 9 ¥ fine, each. 

Relative values, gold to silver as 15^ to 1. 
10 rappen = 1 batzen; 10 b., or 100 centimes, = 

1 franc = 4^ grams of fine silver, - -. = 0.19452 

TRIPOLI (N. Africa).— Tripoli, Sfc.: 100 paras 
= 1 piastre, or ghersch ; value of piastre same 
as that of Tunis. 
TUNIS (N. Africa). — Tunis, Soosa, Cabes, frc. : 
Standard for silver coins = il fine ; for gold coins 
= Yo fine. Relative values, gold to silver as 
15.8125 to 1. 

2 burbine = 1 asper; 52 a., or 16 karob= 1 pias- 
tre =|-| drachma of fine silver, or 44 grains, = 0.123247 

TURKEY. — Constantinople, Smyrna, Aleppo, Trebi- 
zonde, fyc. : Standard for silver coins (assay) == 
T 8 ^ 3 F fine; for gold coins (assay) =-|ff fine. 
Relative values, gold to silver as 15.15625 to 1. 

40 paras, or 100 aspers, == 1 piastre = -^-q checki 

of fine silver, or 15.3856 grains, - - = 0.043096 

1 piastre in gold = 48 1 5o checki of fine gold = 
$0.0437181. 100 piastres in gold = 1 medjdie. 

1 purse of silver — 500 piastres ; 1 purse of gold 
= 30,000 piastres. U. S. Customs value of pi- 
astre = 5 cents. 



16 a FOREIGN MONEYS OF ACCOUNT. 



West Indies. 

CUBA ISLAND. — Havana, Matanzas, Santiago, 
Manzanillo, Baracoa, Cardenas, Cienfuegos, 
Nueviias, Trinidad, fyc : Official as in Spain ; 
also, 12 dineros, or 16 quartos, = 1 real; 8 r., 
or 100 centesimas = 1 duro, peso, piastre, or 
dollar. 

Note. — The full weight Spanish duro colonato is made the unit, or measure 
of value, = $1.04872; but the Spanish onza, or doubloon, passes for 17 dollars, 
and the Mexican and S. American, for 16 dollars. 

HAYTI ISLAND. — Hayti, Port au Prince, Aux 
Cayes, Cape Haytien, Gonaives, fyc. : 
Dominica, San Domingo, Porto Plate, Samana, 8fc. : 
100 centesimas = 1 dollar, or gourda (old Span- 
ish). 

Note. — Spanish doubloons pass for 16 dollars of account; and the Haytian 
silver gourda is worth about | Spanish dollar. 

PORTO RICO ISLAND. — San Juan, Guayama, 
Ponce, Sfc. : Official, as in Spain ; also, 100 cen- 
tesimas = 1 dollar. 
JAMAICA ISLAND. — Kingston, Falmouth, Savana 
la Mar, fyc. : Official, as in Great Britain ; also, 
100 cents = 1 dollar = 1 old peso, or duro of 
Spain = 50 pence sterling in gold, - -=$1.01385 

CARIBBEE ISLANDS. Leeward Islands. — An- 
tigua, — St. John, Falmouth; Dominica, 
montserrat, tortola, vlrgin gorda, 
St. Christopher, Anguilla, Barbuda, 
Nevis, Saba : Same as in Jamaica. 

St. Eustacius : Official, as in Holland. 

Guadeloupe, St. Martin,* Marie Galante, 
Desirade, Les Saintes : Official, as in 
France ; also, 1 00 centimes = 1 dollar ; the 
colonial livre of these islands = -^ standard 
duro colonato of Spain. 

St. Thomas, — Charlotte Amalie ; Santa Cruz, 
St. Jan: Official, as in Denmark; also, 100 
cents = 1 dollar. 

St. Bartholomew: Official, as in Sweden; also, 
1 Spanish dollar = 9 shillings currency. 

* The southern portion of the island of St. Martin is owned and settled by the 
Dutch ; and the moneys of account, weights, and measures of Holland are in 
general use there. 



FOREIGN MONEYS OF ACCOUNT. «17 

Foreign. U. Stales. 

Windward Islands. — Trinidad, Port Spain ; Bar- 
badoes, Bridgetown; Grenada, St. Vincent, 
St. Lucia. Tobago : Same as at Jamaica 
Island. 
Martinique, St. Pierre, Port Royal : Official, as in 
France. 
BAHAMA ISLANDS. — New Providence, Nas- 
sau; Turks, Abaco, Andros, Great Baha- 
ma, &c. : Same as at Jamaica Island. 
LITTLE ANTILLE ISLANDS. — Curacoa,Buen 
Ayre, Oruba : Official, as in Holland. 
Margarita, Tortuga, Blanquilla: Same as 

in V6n6ZllGi£l« 

ZANGUEBAR (E. Africa).— Zanzibar (Island and Town), 
Quiloa, Mombas, Magadoxo, Brava, Socotra Island, fyc. : Accounts 
are now kept in Zanzibar and the Sultan of Muscat's dominions 
on the east coast of Africa generally in dollars of 100 cents ; and 
by the exertions of William E. Hines, Esq., of New York, late 
resident consul of the United States at Zanzibar, the present 
standard dollar in gold of the United States is made the measure 
of value, or, in other words, is officially rated at par. This was 
accomplished in 1865. Other foreign gold and silver coins circu- 
late at conventional rates, some of them above and others below 
their intrinsic values. Thus, the Spanish dollar = $1 ; the Aus- 
trian rix-dollar, scudo, or crown, sells for $1.01 to $1.03 ; the Eng- 
lish sovereign is rated at 4f- Austrian rix-dollars ; the French 
franc in gold, at 18-| cents ; the- silver 5-franc piece, at 94 cents ; 
and the Indian silver rupee, at 47 cents. 

The prices of United-States Bonds, and of American stocks 
generally, -are quoted in Paris in cents per dollar, payable in 
French silver coins, and upon the conventional Bourse rate of 5 
francs to the dollar; whereby the rate of exchange affects the 
prices. From the Paris quotations, therefore, 2.8172 per cent, 
must be deducted to express the true prices, when exchange is at 
par, or when a dollar in United-States gold is quoted at fr. 
5.14086. 

Frankfort. — At Frankfort, the prices of United-States 
Bonds, and of American stocks generally, are quoted in cents per 
dollar, payable either in United-States gold coins, or in guilders 
at the conventional rate of 2^- guilders to the dollar. To these 
quotations, therefore, when payable in guilders, 2.674 per cent, 
must be added to express the true prices. See page 50 a. 
2* 



(18 a) 



FOREIGN LINEAL AND SURFACE MEASURES, 

REDUCED TO THE LINEAL AND SURFACE MEASURES OF THE 
UNITED STATES. 

Foreign. 
ABYSSINIA. — Massudh: 8 robi=l derail, 

or pic, - 
ALGERIA. — 10 decimetres = 1 metre, 

8 robi = 1 pic. Pic, Moorish, for linens, - = 
Pic, Turkish, for silks, &c, - : 
ARABIA. — Muscat : 8 glieria = 1 covid ; 8 
c. = 1 kassaba, - 
500 kassaba = 1 coss, - - -.: 

Aden: 8 robi = 1 yard or pic, 
Jidda : 8 robi = 1 pic, - 

Mocha : 8 robi = 1 gez, - - - 

AUSTRALIA. — Same as in Great Britain. 
AUSTRIA. — Imperial and Legal: 12 zollen = 

1 fuSS, - - - - -: 

29£ zollen = 1 elle, 

6 fusse = 1 klafter ; 10 fusse = 1 ruthe, - ; 
2,400 rutlien rr= 1 meile, - - -: 

192 square ruthen = l metzen ; 3 m. = 1 

jocli, - - * - - -= 1.43 acres. 

AZORE ISLANDS.— Same as in Portugal. 
BALEARIC ISLANDS. — 3 pie, or 4 palmi, 
= 1 vara. 
Majorca: 2 varas = 1 cana, - -=1.711 yards. 

Minorca: 2 varas = 1 cana, - -= 1.754 " 

BELGIUM. — 10 streep=l duim; 10 d. = 1 

palm; 10 p. = 1 el = 1 metre, - - = 1.0936 " 

10 elen = 1 roed ; 100 r. = 1 mijl, or kilo- 
metre, - - = 0.6214 mile. 
100 square roeds = 1 bunder = 100 ares -= 2.471 acres. 
The old Brabant el — 0. 76006 yard. 
BERMUDA ISLANDS. — Same as in Great 

Britain. 
BOURBON ISLAND. — Same as in France. 
CANARY ISLANDS. — Same as in Spain. 
CANDIA ISLAND. — 8 rob = 1 pic, - - = ' 0.697 yard. 

CAPE VERDE ISLANDS. — Same as in Portugal. 



U. States. 


=5 0.682 
= 1.0936 


yard. 

u 


= 0.519 


li 


= 0.692 


a 


= 12.86§ 


feet. 


= 1.2185 


miles. 


= 0.95 
= 0.743 


yard. 

u 


= 0.694 


ii 


= 1.04 


feet. 


= 0.85216 
= 10.39924 


yard, 
feet. 


= 4.7269 


miles. 



FOREIGN LINEAL AND SURFACE MEASURES. a 19 

Foreiqn. U. States. 

CANADA, DOMINION OF. — Same as in 
the United States. 
Lower Canada (special'): 1 arpent, - = 0.8475 acre. 

CAROLINE ISLANDS. — Same as in Spain. 

Central and South America. 

CENTRAL AMERICA. — Costa Rica, Gau- 
temala, Honduras, Nicaragua, San 
Salvador : Same as Spain, — standard 
of Castile. 
SOUTH AMERICA. — Argentine Repub- 
lic : 36 pulgadas = 3 pes = 1 vara, - = 34.1 inches. 
150 varas = 1 cuadra ; 40 c. = 1 legua, -■= 3.229 miles. 
27,000 square varas = 1 suertes des estancia. 
Brazil : 24 pollegadas = 2 pes = 1 covado 

= T 6 o 6 o metre, - - - -=0.7218 yard. 

40 pollegadas = 5 palma == 1 vara = \ 

bracio = 1^ metre, - - -'= 1.203 " 

250 varas, or 125 braccios == 1 estadio ; 8 e. 

= 1 milha ; 3 m. = 1 legoa, - -=4.101 miles. 

4,840 square varas = 1 geira, - -= 1.447 acres. 

Chili: 36 pulgadas = 1 vara (customs'), - = 1.000 yard. 
In all other respects as in Spain, — Castilian 
standard. 
Venezuela, New Grenada, Ecuador, 
Peru, Bolivia, Paraguay, Uruguay : 
Same as Spain, — Castilian standard. 
Guiana : Cayenne : Same as in France. 
Paramaribo : Same as in Holland. 
Georgetown : Same as in Great Britain. 
Falkland Islands : Same as in Great 
Britain. 
CHINA. — 10 tsun, or fan, = 1 punt ; 10 p. = 1 

chih, covid, or cobre (mercers'), - -= 1.226 feet. 

17-i- punts, or 10 tac, = 1 thuoc (tradesmen's) = 0.7152 yard. 
Chih (mathematical), - - -=13.122 inches. 

Chih (engineers' and surveyors'), - -= 12.059 " 

Chih, or kong-pu (architects'), - -=12.709 " 

10 chih = 1 chang or cheung ; 10 chang = 

lyan; 18y.= lli, - - -= 0.3425 mile. 

100 square chih = l mow; 240 m. = 1 fu, 

or king, - - - - -= 0.5564 acre. 

CYPRUS ISLAND.— 8 robi=l pic, -= 0.696 yard. 



20 « FOREIGN LINEAL AND SURFACE MEASURES. 

Foreign. U. States. 

DENMARK. — 24 tomme, or 2 fod, = 1 aln, - = 0.6862 yard. 
10 fod — 1 rode ; 2,400 r. = 1 mill, - = 4.6785 miles. 

280 square rodes = 1 skiepper ; 2s. = l 

toende, - - - - -=1.362 acres. 

6 fod = 1 favn ; and 1 fod = 1 Prussian 
Khein-fuss. 
EGYPT. — 3 kirat = 1 rob ; 8 r. = 1 pic or en- 

drasi (for silks and ivoolens), - -=27.06 inches. 

1 halebi or archin = 27.9 in.; 1 derah = 
25.264 inches. 

2 derah = 1 fedan ; 3 f. = 1 gasab ; 420 g. 

= 1 berri ; 3b. = l parasang, - - == 3.015 miles. 

400 square gasab = 1 fadden al risach, - = 1.465 acres. 
FRANCE. — 100 centimetres, or 10 decimetres, 

= 1 metre, - -=39.37 inches. 

100 metres, or 10 decametres, = 1 hecto- 
metre; 100 hectometres, or 10 kilome- 
tres, = 1 myriametre, - - -= 6.2137 miles. 
100 square metres = 1 are ; 100 a. = 1 hec- 
tare, - - - - -= 2.471 acres. 
1 old aune de Paris = 1.29972 yard; 1 aune 
usuelle, or metrique, = 14 metre. 
FRIENDLY ISLANDS (Feejee and Ton- 
ga Groups). — Same as in England. 
GERMANY. — Prussia, and the Zollverein 
l'angenmaasse for all Germany : 
1 2 linie = 1 zoll ; 1 2 z. = 1 fuss (Rhein- 

fuss), - - - -=12.3514 inches. 

25 1 zollen (Rhein-zollen) = 1 elle = § 

metre, - - - - -=26.24§ " 

10 land-fusse, or 12 Rhein-fusse, = 1 ruthe ; 

2,000 r. = 1 meile, - - -= 4.67855 miles. 

180 square ruthen = 1 morgen, - -=0.6304 acre. 

Special and local : — 

Saxony : 12 zollen = 1 fuss ; 2 f. = 1 elle, - = 0.6196 yard. 

1 Lubec-Brabant elle, - - -= 0.7498 " 

Bavaria : 120 zollen, or 10 fusse, = 1 ruthe, = 9.579 feet. 
34f zollen = 1 elle = | metre, - - = 32.808^ inches. 

Wurtemburg : 100 zollen, or 10 fusse = 1 

ruthe, - - - - -= 9.37326 feet. 

21| zollen = 1 elle, - - -= 0.67175 yard. 

Baden : 20 zollen, or 2 fusse, = 1 elle = -f 

metre, - - - - -=23.622 inches. 



FOREIGN LINEAL AND SURFACE MEASURES. «21 

Foreign. U. States. 

Hesse Darmstadt : 10 zollen = 1 fuss = 

I metre, - - = 0.822^ foot. 

24 zollen = 1 elle = f metre, - - = 0.6561 § yard. 

Mecklenburg : Same as at Hamburg. 

Oldenburg: 1 2 zollen = 1 fuss ; 2f.= l elle, = 0.63528 " 
Bremen : 24 zollen, or 2 fusse,= 1 elle, -=0.63276 " 

1 Bremen-Brabant elle = 1^ Bremen elle. 
Hamburg : 12 zollen = 1 fuss ; 2 f. = 1 elle, = 0.62681 " 
1 Hamburg-Brabant elle, - - -=0.75615 " 

Lubec : 24 zollen, or 2 fusse = 1 elle, - = 0.6294 " 

GREAT BRITAIN. — Same as in the United 

States. 
GREECE. — 60 onue = 5 pes = l passo=lf 
metre. 
22 onue = 1 pichi, for silks, - -=0.701734 " 

23^ onue = 1 pichi, for woolens, &fc, - = 0.749579 " 

HOLLAND. — 10 streep = 1 duim; 10d.= 

1 palm ; 10 p. = 1 el = 1 metre, - = 39.37 inches. 

10 el = 1 roed ; 100 roed = 1 mijl = 1 kilometre. 
100 square roede = 1 bunder = 1 hectare. 
1 old Brabant el - -=0.75931 yard. 

India and Malaysia^ or East Indies. 

Anam : Same as in China. 

Burmah, Pegu : b\ pulgaut = 1 taim ; 4 t. = 

1 sadang; 7 s. = 1 sha, or bambou, -= 154 inches. 

1,000 dhas = 1 dain, - - -=2.4306 miles. 

Ceylon Island : 2 covids = 1 guz, or yard, - = 36 inches. 
Hindostan. — Calcutta : 16 tussoos = 8 gheira 

= 1 haut, or covid; 2 h. = 1 ghes, or guz, = 36 " 

2,000 ghes = 1 coss, - - -= 1.1364 miles. 

20 square covids = 1 chattack; 16 c. = 1 
• cottah; 20 cottahs = 1 biggah, - -= 0.3306 acre. 

Bombay : 16 tussoos = 8 gheira = 1 haut or 
covid ; \\ h. = 1 guz ; 2 hauts, or \\ guz, 
= 1 imperial yard, - - -=36 inches. 

Madras: 16 tussoos = 8 gheira =1 haut or 

covid; 2h.= lguz, - - -=36 " 

Goa : Same as in Portugal. 
Massulipatam : 2 palms = 1 span ; 3 s. = 1 

cubit or covid, - - - -=19^ " 

Swat: 18 tussoos = 1 haut, for matting, -=21 " 

84 tussoos, or 20 wiswusa, = 1 wusa, - = 98 " 



22« FOREIGN LINEAL AND SURFACE MEASURES. 

Foreign. U. States. 

Malabar Coast. — Mangalore, Cananore, Cali- 
cut, Cochin, Quilon, Trivandrum: 3 ghe- 
ria = 1 ady ; 2-| a. — 1 haut ; 2 h, = 1 
guz or guj ah, -. -=38l inches. 

Coromandel Coast, generally, same as at 

Madras. 
Siam. — 48 nions = 4 keubs = 2 soks == 1 ken, = 37.836 " 
2 kens = 1 vouah ; 20 v. = 1 sen ; 4 s. = 

ljod; 25 j. = 1 roeneng, - - = 2.3886 miles. 

Malacca. — Same as at Calcutta. 
Singapore Island. — Same as at Calcutta. 
Philippine Islands. — Same as in Spain, — 

standard of Castile. 
Java Island. — 1 voot (Rhyn-voot), - -= 1.029 feet. 

1 el, or covid (old Amsterdam el), - -= 0.7522 yard. 

English measures also, as at Sumatra. 
Sumatra Island. — 2 tempohs = 1 jancal; 2 

j. = 1 etto ; 2e. = l hailoh, - - = 36 inches. 

ITALY. — The metric system of weights and 
measures is now the official standard, and 
is in general commercial use throughout 
the kingdom. 
Special and local: — 

Milan : 10 atomi = 1 dito ; 10 d. = 1 palmo ; 

10 p. = 1 metro or braccia, - -— 39.37 inches. 

1,000 metri = 1 miglio (kilometre). 
100 metri quadrata = 1 tavolo (are); 100 

t. = 1 tornatura (hectare). 
2^- metri = 1 trabucco ; 4 t. = 1 decametre. 
Florence, Leghorn, Pisa, Sfc. : 2 palmi = 1 

braccia ; 4 b. =1 canna = 2^- metres. 
Carrara : 1 palm for marble = 9.06 U. S. in- 
ches. 
JAPAN. — 10 rin = 1 bun ; 10 b. = 1 tsun ; 10 
t. = 1 sasi or sjak. 
4 tsun-sasi = 5 kani-sasi = 1 hiro ; 5 tsun- 

sasi = 1 ink or tattamy, - - - = 2.07103 yards. 

7 kani-sasi = 1 kan, kian, or ikje, - -= 2.31955 " 

60 inks = 1 tinsc or masti ; 36 t. =± 1 ri, - = 2.54172 miles. 
MADEIRA ISLAND& — Same as in Portugal. 
MAURITIUS ISLAND. — Official as in Great 
Britain ; also, — 
1 aune (old aune de Paris), - - = 46.79 inches. 

MEXICO. — Same as in Spain, standard of Castile. 



FOREIGN LINEAL AND SURFACE MEASURES. a 23 

Foreign. U. States. 

MOROCCO. — 8 pollegadas = 1 palmo da cra- 
veira (Portuguese). 
18f pollegadas, or 2^ palmi da craveira, == 

1 covado or cadee, - - -= 20.21 inches. 

8 rob = 1 pic (Turkish in theory), - = 26.03 " 

NORWAY. — Same as in Denmark. 
NEW ZEALAND ISLANDS.— Same as in 

England. 
PERSIA. — Monlcelser or royal: 12 fingers = 
1 foot ; 1^ f. = 1 cubit ; 2 c. = 1 guz or 
gueza {for silks, §-c), - - -=36.92 " 

1 guz shah (royal), for woolens, - - = 40 " 

1 guz tabree (of Tabreez), - - = 40.4 " 

6,000 guz (monkelser) = 1 parasang, - = 3.496 miles. 
PORTUGAL. — 8 pollegadas = 1 palmo da 
craveira = T 2 ^- metre. 
12 pollegadas = 1 pe ; 2 p. = 1 covado = 

T 6 o 6 o metre - 
40 pollegadas = 1 vara = 1^ metre, - = 43.307 inches. 

24| pollegadas = 1 covado avantejado •(re- 
tail). 

13 a pollegadas = ^ vara = 1 terca. 

60 pollegadas = 1 passo ; 1^- p. = 1 braca. 

780 pes = 1 estado ; 8 e. = 1 milha ; 3 m. 
= 1 legua, - - - - = 3.8386 miles. 

4,840 square varas = 1 geira, - -= 1.4471 acres. 

ROME. — 12 onze = 1 palmo ; l£ p. = 1 pie ; 

5 piedi = 1 passo, - - - = 58.6104 inches. 

10 palmi, or 7 ± piedi, = 1 canna architec- 

tona, - - = 87.9156 " 

3| canne arch. = 1 catena (chain) ; 1 cate- 
na = 10 stajoli. 

8 palmi mercantile = 1 canna mere, -= 78.43932 " 

9 palmi d'are = 1 canna d'are = 1^- metre, = 44. 29^- " 
1,000 passi = 1 miglio, - - -= 0.92504 mile. 
3 % square catene = 1 quartuccio ; 2 q. = 

1 scorzo ; 4 s. = 1 quarta; 4 quarte (7 

pezze) = 1 rubbio, - - - = 4.559 acres. 

RUSSIA. — 6f verschok = 1 foot, - - = 12 inches. 

16 verschok = 1 archine or halebi, -=28 " 

3 archines (7 feet) = 1 sachine, - -= 7 feet. 

500 sachines = 1 verst, - - -= 0.6629 mile. 

2,400 square sachines = 1 deciatene, -= 2.7 acres, 
SANDWICH ISLANDS. — Same as in the 
United States. 



24 a FOREIGN LINEAL AND SURFACE MEASURES. 

Foreign. JJ % States. 

SOCIETY ISLANDS. — Official, as in France. 

SPAIN. — Standard of Castile : 36 pulgadas = 

6 sesma =± 4-quarta or palmos = 3 tercia 

or pie = 1 vara, - - - - = 33.385 inches. 

2 varas = 1 estado, braza, brazada, or 

toesa. 
2 estados = 1 estadale ; 2,000 estadales =' 

1 legua, - - - - - = 4.2153 miles. 

560 square estadales = 1 fanegada, -== 1.592 acres. 

1| pie = 1 codo ; 5 pies = 1 passo. 
Gibraltar : As above ; also, 12 inches = 1 foot ; 

3 f. = 1 yard, - - - - = 1 yard. 

Alicante: 36 pulgadas = 1 vara, - - = 0.8319 " 

Barcelona: 1 vara or matja-c ana, - -- = 0.8641 " 

Santander : 8 octava, or 4 palma, = 1 vara, - = 0.9142 " 
Valencia : 36 pulgadas = 4 palmos = 3 pie 

= 1 vara, - - - - = 1.0044 yards. 

1\ varas = 1 braza-reale. 

Note. — In 1849, Spain legalized the use of the metric system of weights and 
measures in all her dominions; and now, since she is to enforce the employment 
of the money system of France, to the exclusion of all others (see Foreign 
Moneys of Account, Spain), it may be expected that she will unify her weights 
and measures by enforcing the employment of the metric system generally. 

SWEDEN. — 10 linier = 1 turn ; 10 t. = 1 fot ; 

10 f. = 1 st'ang ; 10 s. 1 ref, - - = 97.406 feet. 

20 tumen = 2 fus = 1 aln = \ old aune de 

Paris, - - = 0.64937 yard. 

6 fus = 1 famn; 6,000 f. (3,600 stanger) = 

1 mil, - - - - -=6.6413 miles. 

560 square stanger (14,000 square alner) = 

28 kannland = 16 kappland = 2 spann- 

land = 1 tunnland, - - -=1.21975 acres. 

SWITZERLAND. — Official and general for the 

22 cantons forming the Republic ; — 
10 zollen = 1 fuss; 2 f. = 1 elle = -^ metre, = 23.622 inches. 
4 fusse = 2 ellen = 1 stab = 1 aune metric. 
6 fusse = 1 klafter; 10 fusse = 1 ruthe or 

toise, - - - «- -= 9.8425 feet. 

1,600 ruthen == 1 meile or hour's way = 4|- 

kilometres, - - = -= 2.9826 miles. 

400 square ruthen ss 1 juchart or fekb-aker 

=s 36 ares, - - - -= 0.8896 acre. 

TRIPOLI. — 8 rob = 3 palmi = 1 pic or dra, ^ = 26.42 inches. 
The pic fqr silks = 1 Arabic covid, - - = 19,03, f 



FOREIGN LINEAL AND SURFACE MEASURES. a 25 

Foreign. U. States. 

TUNIS. — 8 robi = 1 pic ; pic for woolens, - = 26.5 inches. 
Pic for silks, - - - - - = 24.83 « 

Pic for linens, - - - .-=18.62 " 

TURKEY. — Constantinople and Smyrna : 1 hal- 
ebi or archin (Russian archine of 28 U. S. 
inches in theory),- - - -=27.9 " 

7,500 halebi = 1 agatch, - - -= 3.3142 miles. 

1 pic, draa, or endrasi, for silks and woolens, = 27.06 inches. 
1 indise, or endese, for cottons, #•<?., - - = 25.688 " 

Bassorah, Bagdad : 8robi = lguz, - - = 31f " 

West Indies. 

In the Islands of Cuba, Hayti, Porto Rico, and Isle of Pines, the 
measures of length and of surface are the same as in Spain, Cas- 
tilian standard, except that in Port au Prince and the French por- 
tion of Hayti, generally, the old piede du roy of 12.78918 U. S. 
inches, and the old aune de Paris of 1.29972 U. S. yards, is used. 

In Jamaica, St. Kitts, Antigua, Montserrat, Tortola, Anguilla, 
Dominica, Barbuda, Nevis, Virgin Gorda; in Trinidad,* Grenada, 
St. Vincent, St. Lucia, Barbadoes, Tobago, Grenadines ; and in 
New Providence, Great Bahama, Turks, t Abaco, and the Bahamas 
generally, the lineal and surface measures are the same as in 
Great Britain and the United States. 

In Gaudeloupe, Martinique, Desirade, Les Saintes, Marie Ga- 
lante, the lineal and surface measures are the metric ; but the pied 
americaine (United-States foot) is chiefly used in measuring tim- 
ber. 

In Santa Cruz, St. Thomas, St. John, the official measures of 
length and of surface are the same as in Denmark, but those of the 
United States are much used. 

In St. Bartholomew : Official, as in Sweden; also as in the Uni- 
ted States. 

In St. Eustatius, Curacoa, Buen Ayer, Oruba : Official, as in 
Holland. % 

In Margarita, Blanquilla, Torluga : Official, as in Venezuela. 
ZANGUEBAR (East Africa) : Same as at Muscat, Arabia. 

* The Spanish (Castilian) vara is still used to some extent by the merchants 
in Trinidad. 



(26 a) 



FOREIGN COMMERCIAL WEIGHTS, 

REDUCED TO THEIR EQUIVALENT VALUES IN THE UNITED 
STATES. 

Foreign. U. States. 

Avoirdupois 
pounds. 

ABYSSINIA (E. Africa).— Massuah : 10 dirhem 
= 1 wakea ; 12 w. or 10 mocha = 1 rotl, ro- 
tolo or litre = 10 troy ounces,* - -= 0.68571 

ALGERIA (Barbary, N. Africa). — Algiers, Bo- 
na, Oron : The metric weights, under French 
denominations, are in official and common use 
here. 
ARABIA. — Mocha : 233^ krat = 1 wakea, wake- 
ga, or vacia= 700 troy grains. 
15 wakea = '1 rotl ; 2 r. = 1 maund or maon, - = 3. — 
10 maund s = 1 frazil ; 15 f.= 1 bahar, - = 450. — 

At the bazaar, 14| frazils = 1 bahar for coffee, — 435. — 
Jidda : Official, as at Alexandria, Egypt ; also, 3f 
drachmas musr (of Egypt) = 1 wakea ; 15 w. 
= 1 rotl ; 5r. = l maund ; 10 m. = 1 frazil ; 
10 f. = 1 bahar = 66§ okes of Egypt, - = 182.8571 

Hodeida, Beit-el-fakih ; 15 wakea = 1 rotl ; 2r.= 

1 maund ; 10 m. = 1 frazil ; 40 f. = 1 bahar, — 81 5.2381 
161 ro tl= 1 tomaun of rice, - - - = 168.14286 

Muscat, Hasek, Sfc. : 233^ krat= 1 wakea; 10 w. 

= 1 rotl ; 9 r. = 1 maund ; 200 m. = 1 bahar, = 1800. — 
<Aden: Official, same as in Great Britain. 
AUSTRALASIA (Oceanica). — Australia, 
New Zealand, Tasmania : Same as in 
Great Britain. 
AUSTRIA {legal for the Empire): 32 lothe=16 # 
unzen = 4 vierdinge = 2 marken = 1 pfund ; 
20 p. = 1 stein ; 5s. = l centner, - - = 123.468 

4 centners = 1 karch ; 5 k. = 1 last ; 2| centners 
= 1 saum ; 1\ centners = 1 lagel; 2 1. = 1 
saum for steel. 
Raqusa, (Dalmatia) : 2£ pfunde= 1 oka, -= 3.0867 

AZO'RE ISLANDS (N. Atlantic Ocean) : Same 
as in Portugal. 

* See Turkey, weights of, and note relative to. 



FOREIGN COMMERCIAL WEIGHTS. «?'/ 

Foreign. U. States. 

Avoirdupois 
pounds. 

BALEARJC ISLANDS (Mediterranean Sea) : 25 

rotoli (22£ Castilian libras) = 1 aroba ; 4 a. 

= 1 quintal, - - - - = 91.3063 

3 quintals = 1 carga; 110 rotoli — 1 oder; 100 

libra nienor = 87 Castilian libra = 1 cantaro 

grosso = 88.2627441 av. lbs. 

BELGIUM. — Same as in France '(metric system). 

BERBER A (E. Africa) : Same as at Mocha, 

BERMUDA ISLANDS (N. Atlantic Ocean) : 
Same as in Great Britain. 

BOURBON ISLAND (Mascerene group, Indian 
Ocean) : The metric system, French nomen- 
clature, is in use here. 

CANADA, DOMINION OF. — Same as in the 
United States. 

CANARY ISLANDS (Atlantic Ocean, W. coast 
N. Africa) : Same as in Spain, Castilian 
standard. 

CANDIA ISLAND, or CRETE (E. Mediterra- 
nean Sea): 100 rottoli = 44 okes=l can- 
taro, - - - - -—116.565 

CAPE OF GOOD HOPE (S. Africa) : Same 
as in Great Britain. 

CAPE VERDE ISLANDS (Atlantic Ocean, near 
W. African coast) : Same as in Portugal. 

Central and South America. 

Guatemala, Honduras, San Salvador, Ni- 
caragua, Costa Rica : 16 onzas == 2 
marcos — 1 libra ; 25 1. = 1 arroba ; 4 a. = 
1 quintal, - -=101.4514 

2£ quintals = 1 carga; 8c. = l tonelada, - — 2029.0285 

Balize. — Official, same as in Great Britain. 

BRAZIL. — 16 oncas= 2 marcos = 1 arratel, -= 1.01187* 

32 arratels = 1 arroba ; 4 a. = 1 quintal, -= 129.5193 

Argentine Republic, or La Plata: 16 oncas 
= 2 marco == 1 libra ; 25 1. = 1 arroba ; 4 a. 
= 1 quintal, - -=101.274 

Note. — The Argentine Republic has established independent standards of 
weights and measures, which are now in practice, and which vary more or less 
in each department from those of Castile. 



28« FOREIGN COMMERCIAL WEIGHTS. 

Foreign. U. States. 

Avoirdupois 
pounds. 

Peru, Chili, Bolivia, Ecuador, New Grana- 
da, Venezuela, Uruguay, Paraguay : 
Generally as in Central America. 
Montevideo : 1 pesado of dry hides (fresh or salt- 
ed) contains 1£ arrobas; 1 pesado of wet salt 
hides contains 2^ arrobas. 

Note. — In Peru, Chili, New Granada, Bolivia, Venezuela, and Surinam, the 
use of the metric system of weights and measures is sanctioned hy law; hut 
as yet (1869) is very little employed in either of the States. 

Guiana. — Cayenne : Same as in France. 
Paramaribo, or Surinam : Same as in Holland ; 

also as in France. 
Georgetown : Same as in Great Britain. 
Falkland Islands. — Same as in Great Britain. 
CHINA. — 10 tseen=l tael or leang; 16 t. = 1 

catty or kan ; 100 c. = 1 pecul or tarn, - = 133.3333 

22f chu = 1 leang ; 2 catties = 1 yin ; 15 y. = 
1 kwan ; 3£ k. = 1 tam ; 1\ t. (60 yin) = 1 
sink, -'-.-'.-- -=160.— 

CORSICA ISLAND (Mediterranean Sea) : Same 
as in France, of which it forms a department. 
CYPRUS ISLAND (Mediterranean Sea): Same 
as in Turkey, and forming a part of Turkey 
in Asia. 
DENMARK. — 32 lod = 16 unze = 2 marken = 1 

pund == £ kilogram ; 100 p. = 1 centner, - = 110.2311 
16 pund = 1 lispund ; 20 1. = 1 shifpund; 16| 

s. = 1 last. 
1 2 punds = 1 bismerpund ; 3 b. = 1 waag or 
vog. 
EGYPT (N. Africa) : 4 gran = 1 kara ; 16 k. = 
1 drachma (oka-drachma) = ^ troy ounce, 
or 48 grains. 
400 drachmas = 4 oka, - - -== 2.74286 

144 drachmas = 1 rottolo ; 100 r. (36 okes) = 

1 cantaro (customs), - - - -= 98.7429 

FRANCE. — 1000 milligrammes = 100 centigrammes 
= 10 decigrammes = 1 gramme == 15.43235 
troy grains. 
1000 grammes = 100 decagrammes =10 hec- 
togrammes = 1 kilogramme, - - = 2.20462 
10 kilogrammes = 1 myriagramme; 100 m. = 

10 quintals = 1 tonneau, - - - = 2204.62143 



FOREIGN COMMERCIAL WEIGHTS. a 29 

Foreign. U. States. 

Avoirdupois 
pounds. 

GERMANY. — Zollverein gewighte for all the 

States of the tariff-alliance : 
10 quentchen = 1 loth; 30 L'=l pfund; 100 

p. = 1 centner = 50 kilograms, - - = 110.2311 

Special and local, or domestic : — 

512 pfennige= 128 quentchene = 32 lothe = 

16 unzen == 2 marken = 1 pfund. 
Prussia : 100 pfunde = 1 centner, - -= 103.1194 

Bavaria : 100 pfunde = 1 centner = 56 kilograms, = 123.4588 
Bremen: 116 pfunde = 1 centner, - - = 127.4887 

Brunswick: 100 pfunde = 1 centner, - -=103.0656 

Hamburg : 112 pfunde = 1 centner, - - == 119.6044 

Hesse Darmstadt: 100 pfunde = 1 centner = 

50 kilos, - - = 110.2311 

Lubec : 112 pfunde = 1 centner, - - -=119.6813 

Mecklenburg : 14 pfunde = 1 liespfund ; 8 1. = 1 

centner, -_._.- -=119.5164 

Oldenburg: 100 pfunde = 1 centner; 3 c. = 1 

pfundschwer, - - - - = 317.7241 

10 pfunde = 1 liespfund ; 29 1. = 1 schiffpfund. 
Wurtemberg : 100 pfunde= 1 centner, . -= 103.1153 

Baden : 10,000 as = 1000 dekas = 100 centas = 10 

zehnling = 1 pfund. 
100 p. = 10 stein = 1 centner = 50 kilograms, -= 110.2311 
Saxony: 10,000 as = 1000 dekas = 100 hektas = 

10 kilas=l pfund; 100 p. = 10 halbstein 

(half-stones) = 1 centner = 50 kilos, - = 110.2311 

Leipsic (domestic) : 100 pfund=l centner, -= 103.0734 

GREAT BRITAIN. — Same as in the United 

States. 

Note. — In Great Britain, in addition to the denominations of -weights used 
in the United States (the values of which are the same), the 

Stone of butchers' meat or flesh. = 8 lbs. 
Stone of cheese, - - - - = 16 " 
Stone of glass, - - - - = 5 " 
Seam of glass, - - - -=120 " 
Stone of hemp, - - - - = 32 " 
Fotber of lead, - -=19£cwt. 

GREECE. — 72 cocos = 1 dramia ; 8 d. = 1 oung- 

hia ; 8 o. = 1 imilitron ; 2 i. = 1 litra, - = 1.136 

400 dramias = 3-|- litras = 1 oka, - - = 3.55 

.13 7£ litras = 44 okas = l cantaro, - - = 136.2 

HOLLAND. — 10,000 korrel=1000 wigtje = 100 

lood = 10 onz = 1 pfund = 1 kilogram, - = 2.20462 

3 * 



Clove of wool, 


-= 7 lbs 


Stone " " iron, flour, 


- = 14 " 


Tod " " - 


-= 28 " 


Weigh 11 " - - 


- = 182 " 


Sack " " - 


- = 364 " 


Last " " - 


-=4368 " 



30« FOREIGN COMMERCIAL WEIGHTS. 

Foreign. U. States. 

Avoirdupois 

HAWAIIAN ISLANDS (Sandwich Islands, Poly- 
nesia, N. Pacific Ocean) : Same as in the 
United States. 



pounds. 



India and Malaysia, or East Indies. 

Annam. — Kesho (Tonquin) : 100 catties — 1 pecul, = 132. — 
Hue, Sai-gon, fyc. (Cochin China) : 16 leang=l 
can ; 10 c. = 1 yen ; 5 y. = 1 binh ; 2 b. = 1 
ta ; 5 t. = 1 quan, - -=688.76 

Buemah (Farther India), Prome, Patanago, Ava, 
fyc: 100 tical orkiat = 8 abucco = 4 agito 
= 3 catties = 1 vis or visay, - - - = 3.39286 

Ceylon Island (Indian Ocean) : 500 pond = 1 

bahar or candy, - - - = 500. — 

Hindostan. — Bombay: 72 tanks =30 pice=l 

maund, - - - - -= 28. — 

Also, 80 tipprees = 40 seers = 1 maund, -= 28. — 

20 maunds = 1 candy = 5 cwt. 
Calcutta, Bengal (factory weight) : 5 siccas or ru- 
pees =1 chattac ; 16c.= l seer; 40 s. = 
1 maund ; 3 m. = 2 cwt., - - - = 224. — 

4 chattacs = 1 pouah ; 5 seers = 1 pussaree. 

I maund (bazaar weight) = 100 troy lbs., nom- 
inally, - - - - -= 82.133 

II factory maunds = 10 bazaar maunds. 
Madras (Carnatic, Coromandel coast) : 10 pago- 
das or varahuns = 1 polam ; 8 p. = 1 seer ; 
5 s. = 1 vis or visay ; 8 v. = 1 maund or 

maon ; 20 m. = 1 candy or baruay, - - = 500. — 

Goa: 32 seers = 1 maund; 20 m. = 1 bahar, - = 495. — 
Pondicherry : 10 varahuns = 1 poloin ; 40 p. = 1 

vis ; 8 v. = 1 maund ; 20 m. = 1 candy, - = 588. — 
Surat : 16 pice= 2 tipprees = 1 seer; 40 s. = 1 

maund; 20 m. = 1 candy, - - -=300. — 

3 candies = 1 bahur. 
Tatta: 4 pice = 1 anna ; 16 a. = 1 seer = 72^ 

tola; 40 s. = 1 maund, - - -= 72.32 

Mysore, Seringapatam, and'Malabar coast general- 
ly : 40 polams = 1 vis or pussaree ; 8 v. = 1 
maund or maon, - - - -= 30. — 

20 maunds = 1 bahur or candy; 20 b. = 1 
garce. 



FOREIGN COMMERCIAL WEIGHTS. a 31 

Foreign. U. States. 

Avoirdupois 
pounds. 
Tranquebar, and Coroniandel coast generally ; 30 
chittacks = 1 vis ; 6§ v. = 1 niaund ; 20 m. 
= 1 candy, - - - - - = 500.— 

Malaya (Malay Peninsula, Strait of Malacca) : 
Same as at Singapore Island ; also, 1 kip for 
tin = 40| Av. lbs., and 20 buncals = 1 catty 
for gold and silver = 2|- lbs., troy. 
Penang Island, or Prince of Wales Island 
(Areca Island, Strait of Malacca) : Same as 
at Singapore. 
Siam (Farther India) : 2 tical = 1 tael ; 20 t. = 1 

catty ; 1 00 c. = 1 pecul, - - -=135.2536 

20 piculs = 1 cajar. 
Singapore Island (off S. extremity of Malay 
Peninsula) : 16 tael = 1 catty; 100 c. = 1 
pecul ; 3 p. = 1 bahar, - - - = 405. — 

Banca Island (Malay Archipelago) : Same as at 

Batavia, Java Island. 
Borneo Island. — Same as at Batavia, Java. 
Celebes Island. — Same as at Batavia. 
Java Island. — Batavia, Sfc: 16 tael = 1 catty; 
100 c. = 1 pecul ; 3 p. = 1 bahur = 200 
goelak, - - - - - = 406.888 

4^ peculs (300 goelaks) = 1 great bahar. 
Moluccas or Spice Islands. — Amboyna, the 
Banda Islands, Batshian, Booro, Ceram, Gi- 
lolo, Oby, Waigeoo : Official, as at Batavia. 
Philippine Islands (Luzon, Mindano, Palawan, 
Mindoro, Panay, Marindique, Negros, Boliol, 
Zuba, Samar, Masbata, Leyte, fyc!) : 100 cat- 
ties = 1 pecul = 37i Castilian libra, - = 139.4957 
1 caban of rice (usual), • - -= 133. — 
1 caban of cocoa, - - - - = 83.5 
Sooloo Islands (Sooloo, Basseelan, Tawee-Tawee, 
Pilas, Pala, Tapul Isles, fyc.) : 10 mace = 1 
tael ; 16 t. = 1 catty ; 50 c. = 1 lachsa ; 2 1. 
= 1 pecul, - -" - - _— i33i_. 
Sumatra Island. — 16 mace = 1 tael ; 25 t. = 
1 catty; 36 c. = 1 maund; 5 J m. = 1 candil 
or bahur, - - = 423.5 
4 catties = 5 goelaks. 1 tael = 133^- lbs., Av. 
14 1 salup = 7| ootan = 1 nelli, 



32 a FOREIGN COMMERCIAL WEIGHTS. 

Foreign. U. States. 

Avoirdupois 
pounds. 
ITALY. — The metric system of weights, either 
under the French denominations or as follows, 
is now the official, and may be considered 
the general commercial system throughout 
Italy, the islands of Sardinia and Sicily in- 
cluded. 
10,000 grani = 1000 denari = 100 grossi = 10 

oncie — 1 libbra = 1 kilogram. 
1000 libbre = 100 rubbi =10 quintali or cen- 
tinaji = 1 migliajo or 1 tonnellata, - - = 2204.6214 

Special and local : 

Carrara : 1 cubic palmo of marble = 884.74 cubic 

inches, - - - - -= 90.82 

JAPAN (N. Pacific Ocean). — NipJwn L, Kioo- 
Sioo I., Sikokf I., the dependencies Yeso I., 
Bonin I., the Loo-Choo group, fyc. : 
1000 moo = 100 rin = 10 sen = 1 monme = 

26.784 troy grains. 
160 monme = 1 kan, - - - -= 0.612206 

350 monme (2 T 3 g kan) = 1 catty; 100 c. = 1 

pecul, - -=133.92 

Note. — In commercial transactions the pecul is usually reckoned at 133| lbs., . 
the same as in China; but it is equal to 133.92 lbs. by the weights and analyses 
of the modern Japanese coins. 

LIBERIA (W. Africa) : Same as in the United 

States. 
MADEIRA ISLANDS (Atlantic Ocean, off W. 

coast of Morocco, N. Africa) : Same as in 

Portugal. 
MALTA ISLAND (Mediterranean Sea, S. of 

Sicily) : Official, same as in Great Britain. 
MAURITIUS ISLAND (Mascarene group, Indian 

Ocean) : Official, as in Great Britain ; also, 

100 livres (old poids de marc of France) = 

1 quintal, - -=107.9184 

MEXICO. — 8 ochavas = 1 onza ; 16 o. = 1 libra ; 

25 1. = 1 arroba; 4 a. =.1 quintal, -=101.4232 

MOROCCO (Barbary, N. Africa) : 10 onza = 1 

mark ; 2 m. = 1 rotl ; 100 r. = 1 cantaro - = 118.65664 

Note. — The commercial rotl of Morocco, both in theory and practice, is equal 
to the weight of 20 old standard duros (silver dollars) of Spain; the miner's rotl 
is equal to 100 meticals (1 old Venetian libbra, peso grosso), or 1.054945 avoirdu- 
pois pound; and the market rotl = 160 meticals. 



FOREIGN COMMERCIAL WEIGHTS. a 33 

Foreign. U. States. 

Avoirdupois 
pounds. 

MOZAMBIQUE (E. Africa) : Same as in Portugal. 

NORWAY. — Same as in Denmark. 

NUBIA (E. Africa) : Same as at Alexandria, 

Egypt. 
PERSIA. — 6 dirhem = 2 mascais = 1 miscal = 
73.846154 troy grains. 
100 miscals = 1 rotl or ratel, - - - = 1.054945 

136^ rotl = 144 avoirdupois pounds. 
6^- rotl = 1 maund tabree (customs), - -= 6.85714 

6 rotl— 1 " " (bazaar), - - = 6.3297 
7% rotl = 1 maund copra (customs), - - = 7.91209 

7 rotl— 1 <•' " (bazaar), - - = 7.3846 
Also, 116f miscals = 1 rotl copra (bazaar), and 

6 rotl copra = 1 maund copra, bazaar. 
2 maunds tabree (of Tabreez) = 1 maund shah 

(of Sheeraz). 
1600 miscals = 1 reh of Teheran = 16 bazaar 

rotl of Tabreez = 6 okes of Turkey, - = 16.88023 

Note. — The maund tabree is used chiefly for weighing coarse metals, coffee, 

sugar, drugs, &c, and the maund copra for weighing rice and provisions. The 

maund shah is used chiefly in Sheeraz, Bushire, and Gombroon, although the 

last-mentioned port now belongs to the Muscat dominion. 

PORTUGAL. — 576 grao = 24 escropulo = 8 ou- 

tava = 1 onca ; 16 o. = 2 marco = 1 arratel, = 1.01187 
32 arratel = 1 arroba ; 4 arroba = 1 quintal, - = 129.5193 
13^- quint alo = 1 tonelada. 
ROME" and Civita Vecchia: 24 grao = 1 oncia ; 12 
o. = 1 libbra ; 10 1. = 1 decime ; 10 d. = 1 
centinajo or cantaro, - - -= 74.7714 

10 centinajo = 1 migliajo. 
RUSSIA.— 96 solotnik =32 loth =12 lana = 1 

funt, = 0.902612 

40 funt = 20 dowinik =13^ trowinik = 8 pa- 

terik = 4 desaterik = 1 pud, - -= 36.10448 

10 puds = 1 berkowitz; 3 b. = 1 paken ; 2 p. 
= 1 last. 
SENEGAMBIA (W. Africa). — Bathurst, Sierre 
Leone : Official, as in Great Britain. 
St. Louis : Official, as in France. 
SOCOTRA ISLAND (Indian Ocean, off E. coast 

of Africa) : Same as at Muscat, Arabia. 
SPAIN. — Standard of Castile: 128 ochavas = 16 

onzas = 2 marcos = 1 libra, - -=1.0145143 



34 a FOREIGN COMMERCIAL WEIGHTS. 

Foreign. U. States. 

Avoirdupois 
pounds. 
25 libras = 1 arroba ; 4 a. = 1 quintal ; 20 q. 

= 1 tonelada, - - - - = 2029.028 

Special and local, but not official : 
Valencia. — Alicante, fyc. : 12 onze = 1 libra me- 
nor (minor). 
18 onze = 1 libra mayor (major) ; 24 1. mayor, 
or 36 1. menor, — 1 arroba =1% Castilian ar- 
roba, - - - - - - = 27.3919 

4 arrobas — 1 quintal ; 2£ q. = 1 carga ; 8 c. 

= 1 tonelada, - - - = 2191.291 

Asturias. — Santander, fyc. : 25 libras = 1 arro- 
ba = li arroba of Castile, - - •' - =± 38.04429 
Aragon. — Saragossa, fyc. : 36 libras == 1 arroba, = 27.3919 
Biscay. — Bilboa, fyc. : 25 libras = 1 arroba = 

l^g Castilian arroba, - - - - = 26.948 

146 libras = 1 quintal macho (for iron). 
Catalonia. — Barcelona, fyc. : 25 libras = 1 ar- 
roba — |- Castilian arroba, - - - = 22.1925 
Andalusia. — Malaga: 7 arrobas (Castilian) = 

If quintal — 1 carga of raisins, - - = 177.54 

Note. — The employment of the metric system of weights is sanctioned by 
law in Spain. 

SWEDEN. — New standard : 100 korn — 1 ort ; 100 

o. = 1 skalpund ; 100 s. = 1 centner, - =± 92.8583 

SWITZERLAND. — New system: 32 lothe, or 8 
gros, = 1 unze ; 16 u. =1 pfund or livre = 
^ kilogram; 100 pfunds = 1 centner, - = 11042311 

TRIPOLI (Barbary, N. Africa.) — 16 karob == 1 

drachma; 10 d. = 1 oncia or usano = 6£ „ 

miscals or 1 troy ounce. 
16 oncia = 1 rotl ; 100 r. = 1 cantaro, - = 109.7143 

400 drachma— 1 oka, - - - - = 2.742857 

TUNIS (Barbary, N. Africa).— Same as in Tripoli. 
TURKEY. — 4 grani = 1 kara, killot, or taim; 16 
k. — 1 dirhem or drachmia = § miscal or 
metical = 49^ troy grains. 
100 dirhems = 1 cheki; 4c. = l oka, - = 2.813187 

250 dirhems = 1 cheki for opium = 166§ mis- 
cals, - - - - - = 1.75836 
Constantinople, Galata : 2 cheki = 1 rotl or rotolo ; 
2 r. = 1 oka ; 50 o. = 1 kantar or cantaro 
grosso, - - - - - = 140.65934 



FOREIGN COMMERCIAL WEIGHTS. « 35 

Foreign. U. States. 

Avoirdupois 
pounds. 
176 dirhems = 1 rotl; 100 r. = 44 okes = 1 

cantaro sottile, - - - -=123.78834 

610 dirhems = 1 teffeh of Brusa silk, - - = 4.29039 

800 dirhems = 1 teffeh of goat's wool. 
Smyrna: 180 dirhems =120 miscals = 1 rotolo; 

100 r. = 45 okes = 1 cantaro, - - = 126.60171 

44 okes — 1 cantaro for tin. 
In Bassorah the Arabs commonly employ the light 
wakea of 160 krat of Mocha = 1 troy ounce, 
or 480 grains ; and 15 wakea = 1 cheki, - = 1.0285143 
40 wakea (2f cheki) = 1 # oka, - - - = 2.742857 

50 okes (2,000 wakea) = 1 cuttra or cantaro, = 137.142857 

Note. — The initial for the avoirdupois values of the Turkish weights, in the 
absence of documentary statistics on the subject, if any exist, was derived from 
the Abyssinian dirhem and by comparison; and the result, I find, is almost 
strictly 'confirmed by assays carefully made at the United-States mint and else- 
where, of the modern gold and silver coins of Ottoman mintage compared with 
their present official standards; viz., -g-i-^ cheki of fine silver to the piastre and 
• 4 g^ cheki of fine gold to the piastre. The troy ounce, it is well known, was 
derived from the Abyssinian dirhem (drachma) or its multiple by 10, the 
wakea, vakia, or wakega, and consists of 12 of the first-mentioned units, making 
the dirhem equivalent to 40 troy or United-States grains, while 120 of these dir- 
hems, or 1 rotl or rotolo of Abyssinia, is equal to 65 miscals, or meticals, or -jL 
maund tabree (customs) of Persia; hence, 120 X 40 —- 65= 7344- troy 
grains, the value of the Persian miscal. But the miscal, or metical of Persia, 
and that of Turkey are alike : in theory it is the same specific weight every- 
where; and 1 dirhem of Turkey is equal to f miscal; hence, 73^Xf = 49y 3 ^- 
troy grains, the value of the Turkish dirhem, and 4 dirhem of Persia are equal 
to 1 dirhem of Turkey, and 6i miscals are equal to 1 troy ounee. 

McCulloch (not to go farther back), in his work published in 1839, says the 
cantaro of Constantinople of 45 okas is equal to 127.2 avoirdupois pounds; or, 
in other words, that the oka of Constantinople is equal to 2.82667 pounds; and 
he states the oka of Smyrna to be equal to 2 lbs. 13 oz. 5 dr., or 2.83203 pounds, 
but, at the same time, under the last-mentioned head (Smyrna), states the 
weights and measures to be the same as those of Constantinople. 

Alexander, in his work published in 1850, places the oka of Constantinople 
at 2.828571 pounds, and that of Smyrna at 2.812488 pounds, in a measure revers- 
ing the values by McCulloch ; while Noback, in a work of more recent date, 
says the oka of Constantinople is equal to 1278.48 grammes = 2.8185644 pounds; 
making it nearly equal to that of Smyrna by Alexander; and that the oka of 
Smyrna is a little heavier, being equal to 2.83286 pounds. 

From these conflicting statements no tenable idea can be gained except this; 
viz., that the initial and leading weights of Asia Minor (Anatolia) are probably 
theoretically and practically the same as those of Turkey in Europe. But this 
seems to admit of no question, since 1 batman of Persian silk, containing 1 reh, 
or 1600 miscals of Teheran, is invariably equal, both in Constantinople and 
Smyrna, to 6 okas of Turkey; wherefore, the oka is equal to ii y ) - Q - = 266|- 

2662. v 73IJL 
miscals, or 3 — i- 3 - = 2|-^ avoirdupois pounds, being slightly heavier 

than that of Smyrna by Alexander, and a trifle lighter than that of Constantino- 
ple by Noback. 



36 a FOREIGN COMMERCIAL WEIGHTS. 

Foreign. U. States. 

West Indies. Avoirdupois 

pounds. 

GREAT ANTILLE ISLANDS. — Cuba : Stan- 
dard of Castile. 16 onzas = 1 libra; 25 1. 
= 1 arroba ; 4 a. = 1 quintal ; 20 q. = 1 ton- 
elada, - . - - - — 2029.028 

Hayti: Poids du marc of France, previous to 

A.D. 1800. 

16 onces = 1 livre ; 100 1. = 1 quintaux ; 10 q. 

= lmillier, - -=1079.176 

2 milliers or barriques = 1 tonneau. 

San Domingo, or Dominica : Same as in Cuba. 

Porto Rico : Same as in Cuba.* 

Jamaica : 16 ounces = 1 pound ; 28 p. = 1 quar- 
ter ; 4 q. = 1 cwt. ; 20 cwt. = 1 ton, - = 2240. — 
LUCAYOS, or BAHAMA ISLANDS. — Same 

as in Jamaica. 
CARIBBEE ISLANDS. Leeward Group. — Do- 
minica, Tortola, Virgin Gorda, St. 
Christopher, Anguilla, Barbuda, Ne- 
vis, Saba : Same as in Jamaica. 

Antigua, Montserrat : 100 pounds = 1 cen- 
tal or cwt., - - - - - = 100. — 

St. Eustacius : Official, as in Holland. 

Guadeloupe, Marie Galante, Desirade, 
Les Saintes : Official, as in France ; also 
as in Hayti. 

St. Martin : Dutch, as in Holland ; French, as 
in France ; also as in Hayti. 

St. Thomas, Santa Cruz, St. Jan: Official, 
as in Denmark. 

St. Bartholomew : Official, as in Sweden. 
Windward Group. — Barbadoes, Grenada, St. 
Vincent, Tobago : Same as in Jamaica. 

Martinique, St. Pierre, St. Lucia : Same as 
in Hayti. 

Trinidad : Same as in Cuba ; official, as in Great 
Britain. 
LITTLE ANTILLES. — Curacoa, Buen Ayre, 
Oruba : Official, as in Holland. 

Margarita, Tortuga, Blanquilla : Same as 
in Venezuela. 
ZANGUEBAR (E. Africa). — Zanzibar (Island 
and Town). Consul's report : 12 maunds = 
1 frasler, - - - - - = 35. — 



(a 87) 



FOREIGN LIQUID MEASURES, 

REDUCED TO THEIR EQUIVALENT VALUES IN THE UNITED 
STATES. 

Foreign. U. States. 

"Wine gallons. 

ABYSSINIA. — By weight: see Weights. 

ALGERIA. — Official, as in France ; also, 16§ litres 

= 1 khoulle ; 6 k. = 1 hectolitre, - - = 26.417 

ARABIA. — (Generally by weight.) Mocha: 20 
wakeas (weight) = 1 nusfiah ; 8n. = l cuda 
or gudda = 16 av. lbs. 

AUSTRALASIA. — Same as in Great Britain. 

AUSTRIA (legal for the Empire) : 4 seidel = 1 
maass ; 10 m. = 1 viertel ; 4v.=l eimer or 
orna, - - - - - - = 14.9543 

32 eimers = 1 fuder ; 42^ maass = 1 eimer for 
beer. 

AZORE ISLANDS.— Same as in Portugal. 

BALEARIC ISLANDS. — Majorca: 8 quartas 
. = 1 quartera ; 3^ quarteras, or 4 quartinel- 
los, = 1 quartin or barril ; 4 quartin = 1 
carga ; 4 c. = 1 botta =27 fluid arrobas, or 
1 pipa of Castile, - - - -=114.9692 

Minorca : 8 quartas = 1 quartera ; 4 quarteras, 
or 2 1 gerah, = 1 quartin or barril ; 4 barrils 
= 1 carga; 4 c. = 1 botta = 3l£ fluid arro- 
bas of Castile, - - - - = 133.1635 

BELGIUM. — Same as in France. 

BERBERA. — Same as at Mocha, Arabia. 

BERMUDA ISLANDS.— Official, as in Great 
Britain ; in trade, generally as in the United 
States. 

BOURBON ISLAND. — Same as in France. 

CANADA, Dominion of. — Official, as in Great 
Britain ; in trade, as in the United States. 

CANARY ISLANDS. — Same as in Spain, Castil- 
ian Standard. 

CANDIA ISLAND. — 1 mistata = 8£ okes weight, 

or, of olive oil, - - - -= 2.9397 

CAPE OF GOOD HOPE. — Same as in Great 
Britain. 

CAPE VERDE ISLANDS. — Same as in Portugal. 
4 



38« FOREIGN LIQUID MEASURES. 

Foreign. JJ. States. 

Central and South America. 

Wine gallons. 

Guatemala, Honduras, San Salvador, Nica- 
ragua, Costa Rica : Same as in Spain, 
standard of Castile ; also the wine gallon of 
the United States is used. 
Balize : Official, as in Great Britain. 
Brazil : 24 quartilhos = 12 garrafa = 6 Canada 
= 3 medida = 1 alqueire or pote =18 arra- 
tels weight, - - - - - = 2.18418 

60 potes = 1 pipa = 1080 arratels weight, - = 131.051 
Bahia : 1 Canada = 15| arratels weight == 5£ 

canadas of Rio Janeiro, - - - = 1.88082 

72 canadas = 1 pipa of spirits,- - - = 135.4193 

100 canadas = 1 pipa of molasses. 
Argentine Republic : 4 cuartos == 1 frasco ; 8 f. 

= 1 caneca = 19 litres in theory, - - = 5.01927 

3 frascos = 1 cortan; 16 c. (6 canecas) = 1 

carga, - . - - - - = 30.11541 

4 cargas = 1 pipa catalana ; also, 8 frascos = 5 
U. S. gallons, nominally. 

Peru, Chili, Bolivia, Ecuador, New Granada, 
Venezuela, Uruguay, Paraguay : Chief- 
ly as in Spain, standard of Castile. 

Note. — In the States last mentioned, the U. S. wine gallon is more or less 
used in trade; and in Chili it is the customs' unit-measure for liquids. Also, in 
Chili, Peru, New Granada, Bolivia, and Venezuela, the use of the metric system 
is sanctioned by law, and may he expected to gradually come into use. 

Guiana — Cayenne : Same as in France. 
Paramaribo (Surinam) : Same as in Holland ; also 

as in Cayenne. 
Georgetown (Demerara) : Same as in Great Britain. 
Falkland Islands : Same as in Great Britain. 
CHINA. — By weight only; for denominations, see 

Dry Measures. 
DENMARK. — 42 pagel = 4 potte = 2 kande = 

1 stubchen, - - - - - = 1.02089 

40 stubchen == 20 viertel == 4 anker = 1 ohm, = 40.83522 
H ohm = 1 oxehoved ; 2 oxehoved = 1 piba ; 

~2 p. = 1 fuder ; 1£ f. == 1 stykfad. 
34 stubchen = 1 toende for beer ; 30 stubchen 
= 1 toende for tar. 
EGYPT. — By weight exclusively. See Weights. 



FOREIGN LIQUID MEASURES. 



«89 



Foreign. 



U. States. 

Wine gallons. 

FRANCE. — 1,000 raillilitres =100 centilitres = 

10 decilitres =1 litre, - - - = 0.26417 

100 litres =10 decalitres = 1 hectolitre, - = 26.417029 

100 hectolitres =10 kilolitres = 1 myrialitre. 
GERMANY. — Prussian and Zollverein maasse for 
all the States of the tariff-alliance : 2 oessel 
•=. 1 quart; 30 q. =: 1 anker; 2 a. — 1 eimer 
= 3840 cubic Rhine zollen ; or, since 38| zol- 
len = 1 metre, = 68§f-|f™£ ii treSj _ _ _ 1&126789 

2 eimers z=. 1 aani, ahm, or ohm ; 3 eimers zzz 
1 oxhoft ; 1 2 eimers zzz 1 fuder. 

3 a eimers = 2 barrile = 1 fass /or beer, - = 60.42263 

Note. — I have been thus particular in treating of the eimer, because the no- 
tion seems to be generally entertained that it is equal in theory to 68.7 litres. 

Special and local, or domestic : 

Baden : 1000 maass = 100 stubchen == 10 ohm=: 

1 fuder =15 hectolitres, - 
Bavaria : 4 quartile =. 1 mass, or masskanne ; 60 
m.r:l eimer, - 

25 eimer z=: 1 fass; 64 masse =1 eimer for beer. 
Bremen : 4 mengel = 1 quartier, or vierling ; 4 q. 
=z 1 stubchen, - 

2\ stubchen =. 1 viertil ; 5v.= l anker ; 4 a. 
= 1 ahm, ----- 
6 ankers = : 1 oxhoft ; 4o. = l fuder ; 44 stub- 
chen =s 1 ahm for wine. 
6 stubchen = 1 steckannen ; 6 s. = 1 tonne for 
train oil = 216 pfunde weight, or, at 7| av. 
lbs to the gallon, - 
Brunswick : 10 stubchen = 1 anker ; 4 a. = 1 
ahm ; 1^ ahm = 1 oxhoft, - - - 

Hamburg : 8 oessel, plank, or stuck = 4 quartier, 
or potts = 2 kannen = 1 stubchen, - 
8 stubchen = 4 viertel = 1 eimer, 
32 stubchen = 1 anker ; 40 stubchen = 1 ohm ; 
60 stubchen = 1 oxhoft ; 240 stubchen = 1 
fuder. 
Hesse Darmstadt : 16 schoppen = 4 masschen 
= I viertel ; 20 v. = 1 ohm =16 decalitres, 
Lubec : 16 ort = 8 plank, or nossel = 4 quartier 

= 2 kanne = 1 stubchen, - - - = 0.9546 

8 stubchen = 4 viertel = 1 eimer, - - = 7.6512 

5 viertels = 1 anker: 6 a. = 1 fass, - - = 57.384 



396.2555 
16.9452 

0.85106 
38.29781 

30.6073 

59.2803 

0.956404 
7.651224 



= 42.26725 



40« FOREIGN LIQUID MEASURES. 

Foreign. U. States. 

Wine gallons. 

Mecklenburg (legal) : Same as in Hamburg. 

Oldenburg : 240 quartiers, or 156 kannies = 1 
oxlioft (legal) = 1 fuder of Lubec. 

Saxony (legal) : 144 nossel= 72 kanne = 24 vier- 

tel=leimer, - - - - = 17.8107 

2 eimers = 1 aam ; 3 eimers = 1 oxhoft ; 5 
eimers = 1 fass ; 1 2 eimers = 1 fuder. 

Wurtemberg. — Helleich mass : 4 quartier, or 
schoppen = 1 maas ; 10 m. = 1 immer ; 16 i. 
= leimer; 6 e. — 1 fuder, - - -=465.9036 

GREAT BRITAIN. — Imperial measure: Denom- 
inations and relative values same as in the 
- United States, but capacity values = 20 ¥ 7 F 4 T 
per cent greater. See Liquid Measures, 
U. S. 
1 imperial gallon =$£Wfr or 1.200320344 wine 
gallons of the United States. 

GREECE. — 1 kila, or galloni = 2^ okas weight. 

HOLLAND. — 10 vingerhoed == 1 maatje ; 10 m. 

= 1 kan; 100 k. = 1 vat = 1 hectolitre, - = 26.417 

HAWAIIAN ISLANDS. — Same as in the United 
States. 

India and Malaysia, or East Indies. 

Annam, Burmah, Calcutta, and Bengal general- 
ly, Ceylon L, Philippine Is., Soo-Loo Is. 
— By weight. See Weights. 
Bombay, Madras : By weight, chiefly ; the wine 
1 gallon of the United States is sometimes used. 
Goa : Same as in Portugal. 
Pondicherry : Official, as in France. 
Malacca. — Capacity measures, same as in the 

United States. 
Penang Island. — Same as in Singapore. 
Siam. — 20 canan == 1 cohi = 80 catties weight, 

or 108.203 av. lbs., - - - - = 12.9757 

Singapore Island. — Capacity measures, same as 

those of the United States. 
Banc a L, Borneo I., Celebes I., Java I., Mo- 
lucca Is., Sumatra I. — Official, same as 
in Holland. 
ITALY. — The metric measures of capacity are 
used here, both under the French nomencla- 



FOREIGN LIQUID MEASURES. a 41 

Foreign. U. States. 

Wine gallons. 

ture and as follows ; viz., 10 coppa = 1 pinta; 
10 p. = 1 mina ; 10 m. = 1 soma = 1 hec- 
tolitre, - - . - -. = 26.417 

JAPAN. — 1 tsjoo = Jg- cubic kani-sasi = 106.09663 

cubic inches, - - - - - = 0.459293 

10 tsjoo = 1 to ; 10 to = 1 kok ; 10 sasi = 1 
goo; 10 goo == 1 tsjoo. 

LIBERIA. — Same as in the United States. 

MADEIRA ISLANDS. — Same as in Portugal. 

MALTA ISLAND. — Official, same as in Great 
Britain ; also, 1 caffiso of oil = 4.724 gallons, 
and 1 barrile of wine = 9.448 gallons. 

MAURITIUS ISLAND. — 8 pintes == 1 velt, - = 1.969 

MEXICO. — Chiefly as in Spain, Castilian stan- 
dard ; but the use of the metric system is le- 
galized, and may be expected soon to be in- 
troduced into practice. 

MOROCCO.— 

MOZAMBIQUE. — Same as in Portugal. 

NORWAY. — The Danish capacity measures are 
used here. 

NUBIA. — By weight, as at Alexandria. 

PERSIA. — By weight. See Weights. 

PORTUGAL. — 24 quartilhos = 6 canadas = 1 

alqueire, or pote = 18 arratels weight, - = 2.18418 

2 potes = 1 almude; 26 a. = 1 bota or pipa, - = 113.5775 

2 botas == 1 tonelada ; 18 almudes = 1 barril. 
Oporto : 1 alqueire = 2 1\ arratels weight, or J g°g°- 

alqueires of Lisbon, - - - - = 3.30936 

ROME, and Civita Vecchia : 64 cartocci =16 

quartucci = 4 foglietti = 1 boccale, - = 0.48165 

32 boccali = 1 barile ; 16 b. = 1 botta, - = 246.605 

32 boccale for wine = 28 boccale for oil. 
RUSSIA. — 100 tscharka = 10 krushka = 1 vedro, 

orwedro, - - - - - = 3.24674 

3 vedros = 1 anker ; 6 a. = 1 oxhoft, - = 58.4413 
40 vedros = 1 botschka. 

SPAIN (Castilian standard) : 4 copas = 1 cuar- 
tillo; 4 c. = 1 azumbra; 8 a. = 1 arroba 
or cantaro = 35 libras weight of distilled 
water at maximum density, or 35.508 av. lbs., = 4.25812 
16 arrobas = 1 moyo; 27 arrobas = 1 pipa; 
30 arrobas = 1 bota ; 60 arrobas = 1 tonelada. 
1 arroba menor/or oil = 27^ libras weight, - ■=: 3.31525 
4 * 



42« FOREIGN LIQUID MEASURES. 

Foreign. U. States. 

Wine gallons. 
Special and local : 

Alicante and Valencia: 16 cuartillos = 4 cuartos 

— 1 arroba = f Castilian arrobas, - -_;== 3.19359 

40 arrobas = 1 pipa ; 2 p. = 1 tonelada, - — 255.4872 

Also 100 cantaros of |^ Castilian arroba each = 
1 tonelada. 
Barcelona: 16 cortans (12 arrobas Catalan 
weight) = 1 carga = 1\ fluid arrobas of 
Castile, - - - - = 31.9359 

Gibraltar : 38 arrobas nienor of Castile = 1 pipa, = 125.9795 
126 U. S. gallons, or 105 imp. gallons in theory 
= 1 pipa. 
Malaga: 33^ fluid arrobas of Castile = 1 pipa, - = 141.9373 

Note. — The employment of the metric capacity measures is sanctioned hy 
law in Spain. 

SWEDEN. — 4 qwarter = 2 stop = 1 kanna — ■£$ 
cubit fot. 
48 kannas = 8 ottingar = 4 fjerding = 1 tun- 

na, - - - - = 33.184106 

SWITZERLAND. — {Official and legal for the 22 
Cantons) : 1000 emine = 100 maass or potts 
=s= 10 gelt = 1 saum = 150 litres, - - = 39.62555 

TRIPOLI. — 14 caraffa= i mataro for oil = 17% 
okas weight, or 48 av. lbs. 
40 caraffa = 1 barril = 50 okas weight ; also 
24 bozza == 1 barile = 130 rotolos weight 
(52 okas), or, of the standard of the United 
States, = 17.10402 

TUNIS. — 2 mettars for wine = 1 mettar for oil = 
36 rotoli weight ; 3^ mettars = 1 millerolle 
= 120 rotoli weight for oil, or 231.65716 av. 
lbs. 

TURKEY. — By weight. See Weights. Also 1 
almud of oil = 8 okas. 

West Indies. 

Cuba, Porto Rico : Same as in Spain, Castilian stan- 
dard ; but in Cuba the U. S. gallon is also 
used : 36 gallons == 1 bocoy = 36 U. S. gal- 
lons. 

Dominica {San Domingo, or Dominican Republic, 
Hayti I.) : Same as in Spain, standard of 
Castile. 



FOREIGN LIQUID MEASURES. a 43 

Foreign. U. States. 

"Wine gallons. 

Hayti, Empire o/(Hayti I.). — 60 gallons = 1 tier- 
con, - - - - - - = 60. — 

Guadeloupe, Martinique, Marie Galante, Les Saints, 
Desirade, northern portion of St. Martin. — 
Official, as in France ; but in trade the United- 
States fluid gallon is chiefly used ; for mo- 
lasses, 30 gallons = 1 baral ; 65 gallons = 
1 tiercon ; 105 gallons = 1 baucaut : for rum, 
114 gallons = 1 boucaut. 

Jamaica, Trinidad, Bahamas, Barbadoes, St. Christo- 
pher, Dominica, Montserrat, Grenada, St. 
Lucia, Antigua, Tortola, Tobago, Nevis, Vir- 
gin Gorda, .Grenadines : Official, as in Great 
Britain ; in trade, mostly as in the United 
States. 

St. Thomas, Santa Cruz, St. Jan : Official, as in 
Denmark. 

St. Eustatius, Curacoa, Buen Ayre, Oruba, southern 
portion of St. Martin : Official, as in Holland. 

Margarita, Tortuga, Blanquilla : Same as in Vene- 
zuela. 



(44 a) 



FOREIGN DRY MEASURES, 

REDUCED TO THEIR EQUIVALENT VALUES IN THE UNITED 
STATES. 

Foreign. U. States. 

Winchester 
bushels. 

ABYSSINIA. — 24 madega = 1 ardeb, - - = 0.33333 

ALGERIA. — Official, as in France; also 16 tarrie 

= 8 saa or saha = 1 caffiso, - - = 9. — 

ARABIA. — By weight : 40 kellas = 1 tomaun 
for rice == 56 maunds weight, or 168 av. lbs. 

AUSTRALIA. — Same as in Great Britain. 

AUSTRIA (legal for the Empire). — 4 becher = 1 

m'assel ; 4 m. = 1 viertel ; 4 v. = 1 metze, - = 1.7452 

AZORE ISLANDS. — 16 quartos = 4 alqueires 
= 1 fanga ; 15 f. = 1 moio =' |- moio of Lis- 
bon, - - - - - - — 20.5298 

BALEARIC ISLANDS. — 6 barcella = 1 quartera 

= 1^ fanega of Castile, - - - = 2.157 

BELGIUM. — 100 kop — 10 schepel = 1 mudde 

= 1 hectolitre, - - - - = 2.83774 

BERBERA. — 

BERMUDA ISLANDS. — Official, as in Great 
Britain ; the U. S. bushel is also used. 

BOURBON ISLAND. — Same as in France. 

CANADA, Dominion of. — Official, as in Great 
Britain ; in trade, as in the United States, ex- 
cept that in Lower Canada the old French 
minot = 1.107436 U. S. bushels is used. 

CANARY ISLANDS. — 12 celamins = 1 fanega 

= 136 libras weight, - - - = 1.77737 

16|- celamins = 1 fanega heaped. 

CANDIA ISLAND.— 1 carga, - - -=4.3211 

CAPE OF GOOD HOPE.— Same as in Great 

Britain; also, 4 schepels=l muid - - = 3.1564 

CAPE VERDE ISLANDS. — Same as in Portugal. 

Central and South America. 

Guatemala, Honduras, San Salvador, Nic- 
aragua, Costa Rica : Same as in Spain, 
Castilian standard ; but the U. S. bushel is 
also used. 



FOREIGN DRY MEASURES. a 45 

Foreign. U. State*. 

"Winchester 
bushels. 
Balize. — Official, as in Great Britain. 
BRAZIL. — 16 quartas = 4 alqueirs = 1 fanga ; 15 

f. = 1 moio = f moio of Portugal, - - = 20.11086 

Bahia : 1 alqueire = 67 ± arratels weight, or 2\ al- 

queires of Portugal, - - - -=0.87985 

Maranham : 1 alqueire = 100 arratels weight, - = 1.30348 
Argentine Republic, Uruguay. — 4 cuartillos 

= 1 fanega = 134 litres, - - - — 3.80257 

Chili. — 1 2 celamins = 1 fanega, - - -= 2.5753 

Peru. — 1 fanega, - - - - -• = 2.31777 

Bolivia, Ecuador, New Granada, Venezuela, 
Paraguay : Chiefly as in Spain, Castilian 
standard. 
Guiana. — Cayenne : Same as in France. 
Paramaribo : Same as in Holland, also as in Cayenne. 

Note. — The use of the metric system is sanctioned hy law in Chili, Peru, 
New Granada, Bolivia, Venezuela, and French and Dutch Guiana, and, to some 
extent, is introduced into practice. 

Georgetown : Same as in Great Britain. 
Falkland Islands. — Same as in Great Britain. 
CHLNA. — 10 ho = 1 shing ; 10 s. = l tau ; 10 t. 
= 1 hwuh, sei, or tane =120 catties weight 
= 160 av. lbs. 
DENMARK. — 32 sextingkar = 16 ottingkar = 2 
skieppe = 1 fjerding, stubchen, or scheffel 
= 36potte, - - - - -= 0.98698 

4 fjerding = 1 toende ; 22 t. = 1 last, - - = 86.8546 

EGYPT. — Cairo : 24 robi = 6 usbek = 1 ardeb = 

144 okas weight, - - - - = 5.088 

Alexandria, Rosetta: 1 kislos= 137 okas weight, = 4.84065 
1 rebeb= 126 okas weight, - - -= 4.45198 

1 ardeb = 230 okas weight, - - -= 8.12663 

FRANCE. — 100 litres =10 decalitres = 1 hecto- 
litre, - - - - - - = 2.83774 

100 hectolitres = 10 kilolitres = 1 myrialitre, - = 283.774 
GERMANY. — Prussia, and Zollverein mass of 
all the States of the tariff-alliance : 16 metzen 
(3072 cubic Rhein zollen, or 48 fluid zollver- 
ein quarts) = 1 scheffel, - - -=1.55776 
Special and local : 
Baden : 1000 becher= 100 m'asslein= 10 sester = 

1 malter = 1$ hectolitre, - - -= 4.25661 

10 malters = 1 zober. 



4G<Z FOREIGN DRY MEASURES. 

Foreign. U. States. 

"Winchester 
bushels. 

Bavaria: 16 dreissiger = 4 massel = 1 viertel, - = 0.525855 
12 viertels (17^ fluid masskanne, or six old inet- 

zcn) = 1 scheffel, - - - - = 6.310263 

Bremen: 16 spirit — 4 viertel = 1 scheffel, - = 2.10289 

10 scheffels = 1 quarter; 4q.= l last, - = 84.11572 

Brunswick : 4 metzen = 1 himt ; 40 h. = 1 wis- 

pel, - - - - - - = 35.3544 

Hamburg : 8 spirit = 2 himten = 1 fass = 1 zoll- 

verein scheffel, - - - - = 1.55776 

10 scheffels = 1 wispel; 6 w. — 1 last, - = 93.4656 

Hesse Darmstadt : 64 kopfchen = 32 maasschen 

= 8 gescheid — 2 kumpf = 1 metze, ' -= 0.45404 
8 metzen = 2 simmer = 1 malter = 128 litres, = 3.632308 
Lubec : 16 fass = 4 scheffels = 1 tonne, - - = 3.9381 

3 tonnen — 1 dromt ; 8 d. = 1 last, - - = 94.5139 
Mecklenburg: 16 spint, or metzen = 4 fass, or 

viertel =1 scheffel, - - - - = 1.103628 

4 scheffels = 1 wispel ; 3 w. = l.last, - - = 105.9483 
Oldenburg: 16 kannen = 1 scheffel; 8 s. = 1 

tonne, - - - . - = 5.17536 

1± tonne = 1 molt; 12rn.= l last, - - = 93.1565 

Saxony: 16 rn'asschen = 4 metzen = 1 viertel, -= 0.73713 
4 viertels = 1 scheffel ; 12 s. = 1 malter, - == 35.3823 

2 malters=l wispel; 6w. = l last, - - =424.5876 

Wurtemberg: 32 viertelein = 8 ecklein = 1 vier- 

ling, - - - . - - = 0.157172 

32 vierling = 8 simri==l scheffel, - -= 5.0295 

GREAT BRITAIN. — Imperial measure : Denom- 
inations and relative values same as in the 
United States, but capacity values = f f -f-fM 
greater; 1 bushel = 1.0315157 U. S. bushels. 
GREECE.— 1 kila, - - - - = 0.944 

HOLLAND. — 1000 maatje = 100 kopen = 10 

schepel = 1 mudde or zac == 1 hectolitre, - = 2.83774 
30 mudden=l last, ... -=85.1322 

India and Malay sia, or East Indies. 

Annam, Burmah, Ceylon Island : By weight. 

See Weights. 
Hindostan. — Bombay : 8 tipprees = 4 seers = 1 

adoulie. 



FOREIGN DRY MEASURES. «47 

Foreign. U. States. 

Winchester 
bushels. 

16 adoulies — 1 para = 8f maunds weiglit, or 

245 av. lbs., - - - - - = 3.15607 

8 paras = 1 candy — 70 maunds weiglit, - = 25.248545 

Calcutta : 80 chattac =16 koonke = 4 raik = 1 
pallie = 5 seers weight, or 9^- av. lbs. : 12 pal- 
lies = 1 morali = 15- factory maunds weight, 
or 112 av. lbs. 
20 pallies = 1 soallee = 2£ factory maunds, or 

186f av. lbs. 
16 soallees (40 maunds, or 2986 f av. lbs.) = 1 

kahoon, - - - - -=38.47397 

Madras : 64 ollock = 8 puddy = 1 marcal. 

5 marcals = 1 para = 5f maunds weight, or 1 35 

av. lbs. 
80 paras= 1 garce = 432 maunds, or 10800 av. 

lbs., - - - - ' - = 139.12464 

Goa : Same as in Portugal. 
Pondicherry : Same as in France. 
Malacca. — The Winchester bushel is used, also 

the coyang of Siam. 
Penang Island. — Generally as in the United 

States. 
Siam. — 40 sat = 1 sesti ; 40 s. = 1 cohi ; 65 c. = 
1 coyang = 52 peculs weight, or 7033.1872 
av. lbs., - - - - - = 90.6009 

Singapore Island. — Generally as in the United 

States. 
Banca Island, Borneo Island, Celebes Is- 
land, Java Island, Molucca Islands, 
Sumatra Island. — Official, as in Holland. 
ITALY. — See Liquid Measures : 

1 soma (hectolitre), - - - - = 2.83774 

JAPAN. — See Liquid Measures : 

1 kok (6i cubic kani-sasi), - - - = 4.93376 

LIBERIA. — Same as in the United States. 
MADEIRA ISLANDS. — Same as in Portugal. 
MALTA ISLAND. — Official, as in Great Britain ; 

also, 1 salma rasa (-f salma colma), - - = 8.2202 

MAURITIUS ISLAND. — Official, as in Great 

Britain.' 
MEXICO. — Chiefly as in Spain, standard of Cas- 
tile ; but the use of the metric system is sanc- 
tioned by law. 



48 a FOREIGN DRY MEASURES. 

Foreign. U. States. 

Winchester 
bushels. 

MOROCCO.— 

MOZAMBIQUE. — Portuguese measures are used here. 
NORWAY. — Same as iu Denmark. 
PERSIA. — 8 sextarios = 2 chenieas = 1 capicha. 
25 capichas = 8 colothuns = 1 artaba = 21 

maunds tabree (customs), or 144 av. lbs., - = 1.8541 
22 sextarios = 1 sabbitha, - - - = 0.20395 

15 capichas = 1 legana, - - - = 1.11246 

PORTUGAL. — 16 quartos = 4 alqueires = 1 fanga 

= 120 arratels weight, - - - = 1.56418 

15 fangas = 1 moio = 1800 arratels weight, - = 23.46267 
Oporto : 1 fanga = 1^ fanga of Lisbon, or 150 ar- 
ratels weight, - - - = 1.95522 
ROME. — 88 quartucci =22 scorzi =16 starelli = 
12 staja = 8 quarterella = 4 quarte = 2 rub- 
biatilli = 1 rubbio, - - - - = 8.3562 
RUSSIA. — 32 garnetz =16 tschetwertka = 4 

tschetwerik = 2 payak = 1 osmin, - - = 2.97607 

2 osmins = 1 tschetwert ; li t. = 1 kuhl. 
SANDWICH ISLANDS. — Same as in the 

United States. 
SPAIN (Standard of Castile). — 16 racion = 4 
quartillos = 1 celemin, or almuda ; 1 2 cela- 
mins = 4 cuartilla = 1 fanega = 4| arrobas 
weight of distilled water at maximum density, 
or 123.64393 av. lbs., - - - = 1.59277 

12 fanegas= 1 cahiz, - -=19.113241 

Special and local : 

Alicante: 4 celamins = l barcella; 12 b. = 1 ca- 
hiz = T \ cahiz of Castile, - - -= 6.950306 
Barcelona: 48 picotin = 12 cortan = l quartera 

= 6| Castilian arrobas weight, - - = 2.08285 

24 quarteras = 1 carga ; 4 quarteras = 1 salma. 
SWEDEN. — 4 quarter = 2 stop = 1 kanna = ^ 
cubic fot. 
7 kanna = 4 kappe=l fjerding, - -= 0.519847 

4 fjerding= 1 spann; 2 s. = 1 tunna, - - = 4.158777 

36 kappe= 1 tunne (firm measure), - - = 4.678624 

SWITZERLAND. — (Official and legal for the 22 
Cantons) : 
100 immi = 10 viertel, gelt, or quarteron = l 

malter= 150 litres, - - - - =4.2566! 

41so, 16 massli = 4 vierling = l viertel. 



FOREIGN DRY MEASURES. a 49 

Foreign. U. States. 

Winchester 
bushels. 
TRIPOLI. = 2 nufs-orbah = 1 orbah ; 4 orbahs = 1 

temen; 4 t. = 1 ueba = 216 rotls weight, - == 3.05279 
TUNIS. — 12 zah, or saha=l quiba; 16 q. = l 

caffiso = 425 okas weight, - - -=15.01663 

TURKEY. — 4 kiloz = 1 fortin = 110 okas weight, = 3.986315 

West Indies. 

Cuba, Porto Rico, San Domingo : Same as in Spain, 
standard of Castile. 

Hayti, Empire of. — 16 litrons = l boisseau; 12 b. 

= lsetiere, - - - - -= 4.4299 

Gaudeloupe, Martinique, Marie Galante, Desirade, 
northern portion of St. Martin, Les Saints. — 
Official, as in France ; also, as in Hayti ; and 
the U. S. bushel is often used. 

Jamaica, Trinidad, Bahamas, Barbadoes, St. Christo- 
pher, Dominica, Montserrat, Grenada, St. Lu- 
cia, Antigua, Tortola, Tobago, Nevis, Virgin 
Gorda, Grenadines. — Official, as in Great 
Britain ; in trade, generally as in the United 
States, but in Trinidad often as in Hayti. 

St. Thomas, Santa Cruz, St. Jan. — Official, as in 
Denmark. 

St. Eustatius, Curacoa, Buen Ayre, Oruba, southern 
portion of St. Martin. — Official, as in Hol- 
land. 

Margarita, Tortuga, Blanquilla. — Same as in Vene- 
zuela. 

St. Bartholomew. — Official, as in Sweden. 



50 # MEMORANDA AND ADDENDA. 



MEMORANDA AND ADDENDA, 

RELATIVE TO FOREIGN MONEYS OF ACCOUNT, COINS, WEIGHTS, 
MEASURES, QUOTATIONS OF STOCKS, ETC. 

. Great Britain. — In Great Britain, sovereigns weighing not 
less than 122f grains are a legal tender for a pound sterling each ; 
which makes the minimum value of a pound sterling in gold equal 
to $4.8458226 ; and this value of the pound sterling, very nearly, 
is adopted by the United-States Government in assessing duties 
on British invoices. 

Goods by weight, passing through a British custom-house, and 
subject to duties, are subject to an allowance, called draft or tret, 
for supposed waste over and above the actual tares, as follows ; viz., 
on 1 cwt. (112 lbs.), 1 lb. ; above 1 cwt. and under 2 cwt., 2 lbs.; 
on 2 cwt. and under 3 cwt., 3 lbs.; on 3 cwt. and under 10 cwt., 
4 lbs; on 10 cwt. and under 18 cwt., 7 lbs; and on 18 cwt. and 
upwards, 9 lbs. These allowances were also made at the United- 
States custom-houses, until July 14, 1862, when the discontinu- 
ance of the practice was ordered by law. These are the chief 
reasons why goods by weight from the United States fall short of 
weight at the British custom-houses. 

Consols, or Consolidated Annuities, represent a considerable 
portion of the public debt of Great Britain : they bear interest at 
the rate of three per cent, a year, payable semi-annually, and are 
transferable. 

The quotations in London of the prices of United-States Bonds, 
and of American Stocks generally, are in cents per dollar, pay- 
able in United-States gold coins ; and upon the old basis of $4-| to 
the £. To these quotations, therefore, 9^- per cent, must be added 
to express the real price. 

France. — The proposed new gold coin of France, of 25 francs, 
to be called an emperor, and to be equal in value to a British sov- 
ereign, must bear a premium at home of about f of 1 per cent., if 
the present mint regulations in other respects be maintained. 

A tolerance of weight in excess of the standard, and in excess 
only, is allowed in France ; while in the United States and in 
Great Britain, strict conformity to the standards are required ; 
thus it happens that the model standards of weight sent abroad 
from France are generally found to be slightly in excess of the 
'true standard. 

The tolerance, or remedy, spoken of is as follows : — 

On iron weights of 50 kilogrammes, tolerance, 20 grammes ; 20 kilogrammes, 
tolerance 10 grammes; 5 kilogrammes, tolerance 4 grammes; 1 kilogramme, tol- 
erance 1 gramme. 

On copper or brass weights of 20 kilogrammes, tolerance U grammes; 5 kilo- 
grammes, tolerance \ gramme; 2 kilogrammes, tolerance J gramme; 1 kilo- 
gramme, tolerance 1£ decigrams. 



CUSTOM-HOUSE ALLOWANCES ON DUTIABLE GOODS. 



«51 



CUSTOMS TARES, 

OR TARES AS ALLOWED BY THE UNITED-STATES GOVERNMENT 
ON DUTIABLE GOODS. 

By an Act of Congress passed July 14, 1862, it is provided 
that when the original invoice is produced at the time of making 
entry thereof, and the tare shall be specified therein, it shall be 
lawful for the collector, if he shall see fit, with Jhe consent of the 
consignees, to estimate the said tare according to such invoice ; 
and that in all other cases the real tare shall be allowed, and may 
be ascertained, under such regulations as the Secretary of the 
Treasury may from time to time prescribe; and that hereafter 
there shall be no allowance for draft. And, in accordance with 
these provisions, the following rates of tare were adopted : — 



Per cent. 


Per cent. 


Almonds, in bags 


2 


Pepper, in single bags 


2 


" in bales 


. 2i 


" in double bags 


4 


" in frails 


. 8 


Pimento, in bags 


2 


Alum, in casks 


. 10 


Raisins, in casks 


12 


" coarse or ground, in 


" in boxes 


25 


sacks, 2 lbs. per sack. 




" in half-boxes 


27 


Barytes . 


. 3 


" in quarter-boxes 


29 


Cheese, in casks or tubs 


. 10 


" in frails 


4 


Cassia, in mats 


. 9 


Rice, in bags . 


2 


Cinnamon, in bales . 


. 6 


Spanish brown, dry, in 




Chicory, in bags 


. 2 


casks .... 


10 


Cocoa, in bags . 


2 


Spanish brown, in oil, in 




" in ceroons 


. 8 


casks .... 


12 


Coffee, Rio, in single bags 


1 


Sugar, in mats 


n 


" " in' double bags 


, 2 


" in bags . 


2 


" all others, actual tare 




" in barrels 


10 


Copperas, in casks . 


10 


" in tierces 


12 


Currants, in casks . 


10 


" in hhds. 


m 


Hemp, Manilla, in bales, ^ 


I 


" in boxes 


14^ 


pounds per bale. 




Salt, fine, in sacks, 3 lbs. 




Hemp, Hamburg, Leghorn 


» 


per sack. 




or Trieste, in bales, 5 lbs 




Tea, China or Japan, in- 




per bale. 




voice tare. 




Indigo, in ceroons 


10 


Tea, all others, actual tare. 




Mel ado .... 


11 


Tobacco, leaf, in bales, 10 




Nails, in bags . 


2 


pounds per bale. 




" in casks 


8 


Tobacco, leaf, in bales, ex- 




Ochre, dry, in cask's 


8 


tra covers, 1 2 pounds per 




" in oil, in casks 


12 


bale. 




Paris white, in casks 


10 


Whiting, common, in 




Peruvian bark, in ceroons 


10 


casks .... 


10 



SECTION B. 

REDUCTIONS, EXCHANGE, INVESTMENTS, MIXED 
NEGOTIATIONS, &c, &c. 

Proposition 1. — When a dollar in gold is worth a dollar and 
thirty cents in currency, what is the value of a currency dollar V 
1-^1.30 = $0.769231, or 76 cents 9.231 mills. Arts. 
Prop. 2. — When gold is nominally at a premium of 3 of per 
cent., at what discount is currency ? 

100 — T ^^ = 26.335175 per cent. Ans. 
Prop. 3. — What is the difference between a given principal, P, 
and a deduction of r per cent, of it ? 

P ~ Pr = P (1 — ?-), the present worth. Ans. 
Prop. 4. — What is the difference between a given principal, 
P, and a discount of r per cent, of it ? . 

Pr P 

P ~ = , the present worth. Ans. 

l_j_ r i_|_ r ' t 

Prop. 5. — Express the difference per dollar between deduction 
at r per cent., and discount at the same rate per cent. 

(1 — r) ~ Ans. 

Prop. 6. — Express the discount, d, on a given principal, P, for 
a given time in days, t, at a given rate, r, per cent, per annum. 

Plr Pi 

d = — = = P — w, i being the interest on 1 

365 -fir l-\-i & 

dollar for the given time at the given rate, and w the present 

365P P 

worth for the same time and rate ; whence, w = == 

365 + *r 1-f-i 
= P — d. See Discount, page 129. 

Prop. 7. — The foregoing proposition (Prop. 6) ; except that 
the time, T, is in years, or years and decimal parts of a year ? 
d = PTr-±-(l-\-Tr). Ans. 
Prop. 8. — The foregoing proposition (Prop. 6), except that the 
time, m, is given in months, or months and decimal parts of a month ? 
d = Pmr -^-( 12 -\-mr). Ans. 
Prop. 9. — To convert time in days into calendar months and 
decimal parts of a month; m = 12d -J- 365 ; but the converse 
of this, viz., d = 365m -f 12, is not liable to be true in practice, 
1 16 



2 b INVESTMENTS, MIXED NEGOTIATIONS, ETC. 

since different calendar months are made up of an unlike number 
of days ; and, therefore, those immediately under consideration may 
contain a greater or less number of days. 

Prop. 10. — What is the intrinsic equivalent in Federal money 
of £ 7 1 2 1 0s. 9 \d. sterling ? 

/ 712 , 12Xs + A x 4.86656 = $3467.6166. Ans. 
\ ^ 12 X 20 / 
Prop. 11. — Reduce $3467.6166 to its intrinsic equivalent in 
sterling money. 

3467.6166 -f- 4.86656 == 712.53958 

20 
10.7916 
12 



9.4992 £712 10s. 9±d. Ans. 
Prop. 12. — Express the equivalent in Federal money, $, of a 
given quantity in sterling money, £, at a given true or direct rate 
of exchange, r, that is to say, at a rate of exchange based upon 
the intrinsic equivalent of the two denominations of money. 
$ = £ X 4.86656 X r. Ans. 
Prop. 13. — Express the equivalent in Federal money of a 
given quantity in sterling money, rated by the former intrinsic par 
of 4s. 6d. sterling to the dollar, the given rate of exchange, f, 
being upon the same basis. 

f _ £ X 40 X/ _ £ X 4.86656 Xf Ans 
9 1.09498 

Example. — What is the equivalent in Federal money of 
£435 7s. 6d. sterling, rated by the former intrinsic par of $4-| to 
the £, the rate of exchange upon the same basis being 1.1 0| ? 
435.375 X 40 X 1.1025 _ 435.375 X 4.86656 X 1.1025 _ 
9 1.09498 ~ 

$2133.3375. Ans. 
Prop. 14. —Reduce Fll 72.36 (1172 francs, 36 centimes of 
France) to its equivalent in Federal money ? 

1172.36 X 0.19452 = 1172.36-^-5.14086 = $228.05. Ans. 
Prop. 15. — Express the intrinsic or par equivalent of francs, 
F, or francs and decimal parts of a franc, in Federal money, $ ; 
and vice versa. 

$ = F X 0.19452 = F-f- 5.14086. Ans. 
F = $~ 0.19452 = $ X 5.14086. Ans. 
Prop. 16. — Express the intrinsic equivalent of marks banco, 
M, or marks and decimal parts of a mark banco of Hamburg, 
London rate, in Federal money ; and vice versa. 

$ = M X 0.35393 = M -f- 2.8254. Ans. 
M= $ X 2.8254 = $ -f- 0.35393. Ans. 



INVESTMENTS, MIXED NEGOTIATIONS, ETC. 53 

Example. — What is the equivalent in Federal money of 6472 
marks, 1 2 schillings banco of Hamburg, London rate V 

6472.75 X 0.35393 = 6472.75 -f- 2.8254 = $2290.90. Ans. 
Prop. 17. — Which is the most advantageous purchase, other 
things being equal; viz., bills of exchange on London at 1.10£, 
rate of 45. Sd. sterling to the dollar; or on Paris at F5.15 to the 
dollar; or on Hamburg at 35| cents per mark banco ? 
1.105 -^ 1.09498 = 1.009151- London. 
5.15 -f- 5.14086 = 1.00178- Paris. 
35.625 -f- 35.393 = 1.00655+, Hamburg. 

Bills on Paris. Ans. 

Note. — London allows, as the intrinsic par, £1 for F25. 21 =19.304 Federal 
cents per franc, instead of 19.452, the prevailing commercial pur. But at the 
same time G-reat Britain values the silver franc of France, measured by the sil- 
ver in her own silver coinage, at £1 = ~f'9-'4~tlj"f 73"^ ~ 23 - 2i9 "l~ francs = 
20.93234+ Federal cents per franc. 

Of Notes for discount, and their avails at bank. 
Prop. 18. — When the time of the note is expressly written in 
days-; or when the actual number of days in the specified time are 
employed ; that is, when the true time is taken, and taken in days, 

P = , and a = -A_ — — — L P being the principal or 

365 — tr 365 oil 

face of the note, a the avails or sum advanced by the bank, t the 

time including grace, and r the rate of the interest or discount per 

annum. See Bank Discount, p. 127. 

Example. — What must be the principal of a note payable in 

90 days, in order that the equitable avails at bank, for the time of 

the note and 3 days grace, the rate being 6 per cent., shall be 

$1000? 

365 X 1000 tt-.n, K KO a 

— =$1015.53. Ans. 

365 — 93 X -06 
Prop. 19. — When the time of the note is expressly written in 
months, and a month is taken to be the -^ part of a calendar 
year, 

r> 12a , P[12 — r(w X -1)1 i. • 

P = — . , and a = —h i — _ — £J ; m being 

12— r(m-L-0.1) 12 

the time in months, and . 1 being -^ of a month, or the usual 3 
days grace. 

Example. — $1000 are to be obtained from a bank, on a note 
having three months to maturity, and three days grace ; the rate 
being 6 per cent., for what sum must the note be drawn ? 
12X100 =1015 . 74 . Ans , 
12—3.1 X -06 

Note. — 3 calendar months and 3 days, mean time, are more than 93 days, 
by 1£ day. 



45 INVESTMENTS, MIXED NEGOTIATIONS, ETC. 

Prop. 20. — When the true time of the note is taken in months 
or days, and a year is assumed to consist of 12 months of 30 days 
each, or of 360 days only, 

P = . 36ha , and a = P V 60 ^1. 
360 — tr 360 

Example. — What must be the principal of a note for discount, 
payable in 90 days after acceptance, that the proceeds at bank shall 
be $1000, the rate being 6 per cent., the grace 3 days, and the 
bank assuming that a year consists of 360 days only ? 
360X1000 =S101 5.74, Ans. 
360 — 93 X -06 
Note. — The Government of the United States, and the Courts, in matters of 
interest and discount, reckon time at 365 days to the year; and in Great Britain, 
France, and all Europe, a year, for like purposes, contains 365 days. 

Prop. 21. — A note on time, without interest, dated Jan. 2, 

1868, is to be given in exchange for the following obligations ; the 
time of the note is required. 

Note, due March 3, 1868, for $370. 
Bond, " April 16, " 830.50 

Note, " June 11, " 1120. 

Acc't, " June 4, " 127.50 

From Jan. 2 to March 3 is 61 days X 370 = 22570 
" " 2 to April 16 is 105 " X 831= 87255 
" " 2 to June 11 is 161 " X H20 = 180320 
" " 2 to " 4 is 154 " X 127 == 19558 

2448 )309703(127 days 
after Jan. 2 = May 8, 1868. Ans. 

But this is simply equivalent to finding the common time of 
maturity of the obligations to be transferred, which must be the 
answer. 

Example.— Due March 3, $370 X = 

" April 16, 831 X 44 = 36564 
" June 11, 1120 X 100 = 112000 
" " 4, 127 X 93 = 11811 

2448 ) 160375(66 days 

after March 3, = May 8, 1868. See Equation of Payments, 
page 132. 

Prop. 22. — A note dated March 1, 1869, and bearing interest 
at 7 per cent, from date, is to be given in exchange for the follow- 
ing obligations ; the principal of the note is required. 

1869, Feb. 16, settlement note on interest, value of, this 

day, $327.36 

u March 27, acceptance of this date for 1000.00 

" May 4, account, averaging due this date, 658.73 



INVESTMENTS, MIXED NEGOTIATIONS, ETC. 55 

1869, Feb. 16, due $327 

" M'ch. 27, " 1000 X 39 = 39000 
« May 4, " 659X77 = 50743 

$1986 )89743(45.2 days later than 

Feb. 16, = April 3, 1869, the day on which the given obligations 
collectively become due, or average due; which is 32 days later or 
after March 1, 1869 : then, since the adjustment involves discount 
instead of interest, 

p = JM*_ = 365X1986 =<1978>8 ^ AfUm 
365 -\-tr 365-}- 32 X. 07 
In which P represents the principal, or face of the new note, a 
the sum of the obligations to be transferred or cancelled, t the time 
in days of the adjusting interest or discount, and r the rate of the 
interest or discount. 

Prop. 23. — The foregoing proposition (Prop. 22), except that 
the new note is to be dated July 1, 1869, instead of March 1, 1869. 
From April 3, '69, to July 1, '69, is 59 days, the number of days 
that April 3d is earlier or previous to July 1st : then, since the ad- 
justment involves interest instead of discount, 

p = fl (865 + fr) = 869.18 X 1»86 =$ y> ^ 

365 365 

Prop. 24. — Suppose we equate the time of the following de- 
mands by the common rule, and then make up the account correct- 
ly by interest and discount to the given dates, and thence to the 
equated time, by way of exhibiting general principles, and the 
bearing of the common rule of average upon those principles. 
Interest and discount at 7 per cent. 
1868, note due March 10 for $500 

" " " June 8 " 750 X 90= 67500 
" " " Sept. 6 " 1050 X 180 = 189000 

2300 ) 256500 ( 112 days 

later than March 10, = June 30, '68, the day on which the given 
demands ($2300) are assumed to collectively mature. 

$500 due March 10, worth March 10 $500. 

750 " June 8, " " 10 (90 days' disc't.) 737.28 

1050 " Sept. 6, " " 10 (180 " " ) 1014.96 

$2252.24 ; 
worth, June 30 (112 days' interest), $2300.62. (True time = 
365 (P — w) -^ wr = 112 — 365 {W— P) -f- Pr.) 

$500 due March 10, worth June 8 (90 days' interest) $508.63 
750 " June 8, " " 8 750.00 

1050 " Sept. 6, " " 8 (90 days' discount) 1032.18 
x * $2290.81 ; 



6 b INVESTMENTS, MIXED NEGOTIATIONS, ETC. 

worth, June 30 (22 days' interest) $2300.48. (True time = 
365(P — w) -^ wr -\- d0.) 

$500 due March 10, worth Sept. 6 (180 days' interest) $517.26 

750 " June 8, " " 6 ( 90 " " ) 762.95 

•1050 " Sept. 6, " " 6 1050.00 

$2330.21 ; 
worth, June 30 (68 days' discount), $2300.21. (True time = 
180 — 365(iv — P)-±-Pr.) 

$500 due March 10, worth June 30 (112 days' int.) $510.71 
750 " June 8, " " 30 ( 22 " " ) 753.17 
1050 " Sept. 6, " " 30(68 " disc't.) 1036.48 = 
$2300.36. (True time = 112 — 365(w — P) -f- Pr — 111.184 
'days. 

Note. — On the assumption that simple interest is justly due and payable 
yearly, it is apparent that no demand should enter the account for average at a 
present -worth more than a year due previous to the maturity of the debt latest 
due. 

Prop. 25. — A man sold two horses for $150 apiece, one at a 
profit of 25 per cent., and the other at a loss of 20 per cent. ; 
which was the greater, the profit or the loss on the two sales, and 
what sum of money expresses the difference ? 

150 -j- (1 — .20) =$187.50, cost of the horse sold at a loss, and 
187.50 — 150 = $37.50 loss ; 150 -^ (1 + .25) = 120, cost of the 
horse sold at a profit, and 150 — 120 = $30 profit ; then 37.50 — 
30 = $7.50 loss greater than the profit. Ans. 

Prop. 26. — A merchant sold two packages of goods for the 
same sum of money each ; on one of them he cleared 25 per cent., 
and on the other he lost 25 per cent. : did he gain or lose in the 
aggregate ; and, if either, what per cent. ? 

1 -j- (1 — .25) = 1|-, cost relative to the sum received as 1 of 
the package sold at a loss ; 1 -f- (1 -J- .25) = T 8 „, cost relative to 
the sum received as 1 of the package sold at a profit ; then 
(11—1)^(1— -fa) _ _ 4 ^ = _j_ _ 6 i per cent logg Ang ^ 

I3- -\~ iiy 

Prop. 27. — A and B purchased a farm in company for $8000 ; 
A paid $5000 in part payment, and B paid the remainder ; they 
then sold ^ of the farm for $4000 ; and, to close the copartnership, 
each took \ of the remaining § to his own private interest : how 
much, if any, of the undivided cash received for the land they 
disposed of attaches to B's private interest ? 

|(4000 — \ of 8000) = $500, B's share of the profit on the 
sale, and 3000 -4- 500 — \ of 8000 = $833£. Ans. 

Prop. 28. — The sum, S, of two numbers; N and n, given, the 
difference, d, of their respective factors, C and c, to produce like 



INVESTMENTS, MIXED NEGOTIATIONS, ETC. 



b7 



products given, and the sum of the products, P, given, to find the 
numbers. 

Let D — the difference of the assumed numbers, and letwi = 



1st step, *S(m + jd) = N iP_ 

m 

S — N=n 
N — n = D 

d' : d : : 

2d step, gib*? = N' 



C — c = d', and 



S — N' = n' 



±P-±N' = c, C' — c' — d' 



3d step, 




: D' : D" 

%P-^-n"=C" 

xp 4- i\r" = c", and C 

' = d"' &c. 



Note. — By this manner of proceeding, all the elements in propositions of this 
nature may he approximated to any degree of exactness desired; and the true 
values will he between those obtained by the first step and those obtained by the 
second. The greater required number (A r ) will be less than that obtained by 
the first step, and greater than that obtained by the second. Often but two steps, 
and seldom more than three, will be required for ordinary practical purposes. 
When the difference between any trial C and its corresponding c becomes equal 
to the given difference (rf), the elements in that step will be the exact ones sought, 
if no error has been committed in the work. It is not necessary, however, to ob- 
tain the first trial N by the method here proposed; for any number whatever 
may be assumed in its stead : but, when thus obtained, it has the advantage, gen- 
erally, of being near the true number sought, and of being known to be greater 
than that number. 

Example. — A certain farm containing 80 acres is worth in 
the aggregate $62.50 per acre ; but one section of it, to the extent 
of half the gross value, is worth $11 per acre more than the other ; 
how many acres are there in the lesser section ? 

P = 80 X 62.50 = $5000, the gross value of the farm ; then 



= 43.52r=iV 



( 62.5-J-5. 5) 40 . 

62~X 
80 — 43.52 = 36.48 = n 



2500 



36.48 = 68.5307+ =C 



2500 -±- 43.52 = 57.444853- — c 
= d> 



11.085846 
6.985484+ z=D' 



7.04 = D 

n.085846 : ii :: 7.04 

80 + 6.985484 =43>492742 = iV/ 

2 
80 — 43.492742 = 36.507258 = n' 

2500 -4- 36.507258 = 68.479534 = C 
2500 + 43.492742 = 57.480855 = c 

10.998679 =d" = 
11 nearly; the lesser section, therefore, contains 36.507258 acres, 
nearly. Ans. 



8 b INVESTMENTS, MIXED NEGOTIATIONS, ETC. 

Note. — All propositions of this general class having given relations of parts 
contain the requisite elements for direct solutions, and need not be worked, as 
they commonly are, by rules denominated Double Position. For example : 
" The old sea-serpent's head is 10 feet long, his tail is as long as his head and 
half the length of his body, and his body is as long as his head and tail; what 
is the whole length of the monster ? " 
h = 10,t = h + ^b,andb = 2Ji+{b; then 

b — %b = 2h, therefore b = 4h = 40 
t = h + \b , therefore t = 10 -f- 20 = 30 

h = 10. 80 feet. Ans. 

Prop. 29. — Divide $1000 into four such parts that the second 
shall contain $10 more than the first, the third $6 more than the 
second, and the fourth 2£ times as many as the first and second. 
Let x represent the smallest part ; then 
1000 = a? -|- a; + 10 + a: +16 + 5a: +25, and 
1000 — (10 + 16 + 25) = 949 = 8a:, and 949 4- 8 = x = 

$118,625, 1st; $128,625, 2d; $134,625, 3d; $618,125, 4th. 
Prop. 30. — A gentleman being asked his own age and the age 
of his wife, replied, If you subtract 5 years from my age, and di- 
vide the remainder by 8, the quotient will be £ of my wife's age ; 
and if you add 2 years to her age, then multiply the sum by 3, and 
subtract 7 from the product, the remainder. will be my age. Ke- 
quired the age of each. 

Let x = the wife's age ; then 3a: + 6 — 7 == the husband's age ; 
but 3a: + 6 — 7 — 5 _ x therefore 9x i 18 _ 36 _ $ x an & 

8 — 3' ^ 

x = 36 — 18 = 18, the wife's age, and 18X3 + 6 — 7 = 53, the 
husband's age. Ans. 

Prop. 31. — Two relations between two unknown numbers given, 
to find the numbers. 
By Double Position. — Special cases. 
Let x = the lesser of the correct numbers sought. 
" N = the greater and n the lesser of the assumed values of 

x, or trial numbers. 
" D — the greater and d the lesser of the differences between 
the trial numbers and those obtained in their stead ; then 

When both differences are in excess, or show that both the assumed 
values of x are too high, 

(N+n)d 
D—d 
When both differences are in deficiency, or show that both the as- 
sumed values of x are too low, 

(N+n)D 

D+d" 

When one of the differences is in excess and the other in deficiency, 



INVESTMENTS, MIXED NEGOTIATIONS, ETC. 59 

or when one of the assumed values of x is too high and the other too 
low, 

_ND + nd — D—d 

x £q^ ' 

Example. — Assume the wife's age (last preceding proposition) 
is 25, then (25 + 2) X 3 — 7 = 74, the husband's age, and 

(74 5) X3 __ 25.875, instead of 25, and showing a difference 

8 
of .875 in excess. Again, assume the wife's age is 20, then 

(20 + 2)3 — 7 = 59, the husband's age, and ( 59 ~ 5 ) 3 = 20.25, 

8 
instead of 20, and showing a difference of .25 in excess ; then 
(25 + 20) X -25 _ lg the ife , s " and 
.875 — .25 
(18 + 2) X 3 — 7 = 53, the husband's age. 
Proof. (53 — 5)3 =18 _ 
8 
Note. — It is not necessary to assume the value of the lesser required num- 
her instead of that of the greater, for the value of either may he assumed as 
preferred ; and when the greater is assumed, the foregoing formulas are not ap- 
plicable, hut a set that are can he easily made. 

Prop. 32. — A said to B, Give me one of your apples and I 
shall then have as many as you will have left. B replied, Give me 
one of yours, and I shall then have twice as many as you will 
have left. How many apples had each ? 

By the first proposition, A-\-l = B — 1 ; therefore B = A-\-2: 
but by the second proposition, B + 1 — 2(A — 1) ; therefore B = 
2A — 3 : then 2A = A + 2 + 3, and A= 2 + 3=5; B = 
A + 2 == 7. Ans. 

Prop. 33. — The lesser and half the greater of two casks of 
wine =82 gallons; and the greater and ^ the lesser = 129 gal- 
lons : how many gallons are in each cask ? 

Let x = the greater and y the lesser ; then 
x + ^y = 129, and y + \x = 82 : therefore 
2y= 164 — x, and %y ■=. 129 — x\ consequently 
2^—^=164—129, and # = 21 ; also 
129 — &=-£== 129— 7=122. 'Ans. 
Prop. 34. — Smith's several cows are in number to their aver- 
age cost per head as 45 to 138, and they collectively cost him $690 ; 
how many cows has he ? 

Let n represent the number of cows, and c their average cost ; then 
n : c : : 45 : 138 ; but n X c = 690, therefore 



J 



690 X 45 1K A 

— — - = n = 15. Ans. 

138 

Solved, also, by Prop.. X, page 1 70. 



x = 



10b INVESTMENTS, MIXED NEGOTIATIONS, ETC. 

Prop. 35. — An apple-vender bought one-half of a certain lot 
of apples at the rate of 2 for a cent, and the other half at the 
rate of 3 for a cent, and concluded that they collectively cost him 
2 cents for 5 ; being willing to dispose of them at cost, he accord- 
ingly mixed them together, and sold them out 5 for 2 cents, and 
lost 5 cents by so doing : how many apples had he ? 

Let x = the whole number of apples or answer ; then 

/j? + £ Aj /x + ; A5 = 5to-600 therefore 
\,2 ^3 /2 \4 *6 /2 48 

48 x -4- 600 = 50a:, 2x = 600, and x = 300. Ans. 
Note. — The sum of £ of a quantity and of | of a like quantity, is more than* 
■| of the sum of the two quantities, hy -g 1 ^- of one of the quantities. 

Prop. 36. — If a body were to start suddenly into motion, and 
move 80 miles in the first hour, and in each succeeding hour were 
to move through three-fourths as much space as in the hour last 
preceding, and were thus to continue in motion forever, whajt space 
would it describe ? 

80 -*- 1 = 320 miles. Ans. 

Note. — This is simply a question in Geometrical Progression descending, in 

which the greater extreme is 80, the ratio £, the less extreme 0, and the sum of 

the terms is required. The formula, since the ratio is less than unity, becomes 

g _ (E--e)_(jL — r) + E _ ^Xl-g, gee Geometrical Progression, p. 151. 

r — e r — e 

Prop. 37. — What sum in ready money, D, may I pay for 
Si 0,000 in stocks, P, that are redeemable at par in T, 3 years, 
and are bearing interest the while at 7 per cent, a year, payable 
half-yearly, in order that I may realize 6 per cent, simple interest 
a year on the investment, supposing that the payments of the in- 
terest on the stock are to be kept invested at 7 per cent, a year, 
from their times of maturity till the stock matures ? 

A =P+ P T(1 + *<£=» + iB)=p[l+« (l+ r -^)], 

the amount of the stock at the time of its maturity ; in which p 
represents an interest payment, r the rate of the interest per in- 
terval between the payments, and n the whole number of the in- 
terest payments; and D = A -"- (1 -^-R'T) = A — (1 -j-an), in 
which R* represents the rate of the discount per annum, or rate 
of the interest on the investment per annum, and a the rate of the 
discount per interval between the payments ; therefore 
10000 -|- 700 X 3 X (1 + ^^|=il + '¥) _ 
1.18 
10000 X [1 + -035 X 6 X (1 + ^-^ & = n )'\ _ 
l7l8 

$10,409.96. Ans. 



INVESTMENTS, MIXED NEGOTIATIONS, ETC. 611 

Prop. 38. — The last preceding proposition, except that my 
ready money is worth 6 per cent, compound interest yearly ? 
D = A^-(l-\-Rf=z 12283.75 4-1. 191016 = $10,313. 67. Ans. 

See Annuities, p. 156 ; also, p. 125. 
Prop. 39. — The preceding proposition (Prop. 37), except that 
the payments of the interest are to be invested at 6 per cent, a 
year, from the times they become due, till the stock matures ? 
10000 x (1 + -03 X 6(1 + *V*) = >10>114 . 41 . Ans . 
1.18 
Prop. 40. — The preceding proposition (Prop. 37), except that 
the interest on the stock is payable quarterly ? 
10000 -|- 700 X 3X (l+ - 07x 2 f3 ~ 1) +|X-07) _ 
1.18 
10000 X (1 + .0175 X 12X(l+ ljLLI sf- u ) _ 
1.18 

$10,427.22. Ans. 
Prop. 41. — The first proposition in this class (Prop. 37), ex- 
cept that my ready money is worth 6 per cent, interest a year, 
with the interest payable semi-annually ? 
D _ A _ A _ 12283.75 _ 

— \ + T{R' + lR') ~ 1 -1- TQ^-Y-) ^225 

$10,027.55. Ans. 

Prop. 42. — Express the difference, per dollar, between the 

amount of a given algebraic principal for a given algebraic time 

and rate, and the present worth of the same principal for the same 

time and rate, the time being in days. 

365 + tr 365 A . tr \ 1 A 

_ ! — ^ = ( 1 4- — ) ^ . Ans. 

365 365 + tr \ ' 365/ 1 -f- gf T 

Example. — What is the difference between the amount of 
$1,550 for 175 days at 8 per cent, a year, and the present worth 
of the same sum for the same time and rate ? 

^_ P(365-ffr) 365P _p/365+*r_ 365 \_ 

365 ~~ 365 -|- tr ~ \ 365 365 -j- tr) ~~ 

$116,708. Ans. 
Prop. 43. — A purchased a bill of goods on six months' credit, 
amounting to $2,000, with the understanding that he should be 
allowed 5 per cent, off for ready cash, in whole or part payment : 
he paid $1,000 ready cash ; for what sum ought he to be credited 
on the bill ? 

1000 -r-(l — .05) =$1052.63. Ans. 
Prop. 44. — Which is the lower offer, goods at 1.28f, on 4 
months' credit ; or the same goods at 1.30, on 6 months' credit 5 
allowing money to be worth 8 per cent, interest a year ? 



12b INVESTMENTS, MIXED NEGOTIATIONS, ETC. 

1.28| 1.2875 X 12 , nKA +T . ,, 
-^_ = ~ — = 1.254+, the present worth. 

l+AXg-- 12 - 32 

1.30 1.30 X 

f+^P 12.48 



1.30 1.30 X 12 - nK . T . ^ 

. — = 1.25, the present worth. 



The 1.30 terms, slightly. Ans. 



Conversion of debts not yet due into others of like sums 
each, and having a common difference of time from ma- 
turity to maturity. 
S = gross sum to be converted. 

T = time from the present to the maturity of the gross sum. 
n — number of common substitutes. 
s = common sum of the substitutes. 
t = common difference of time from maturity to maturity of the 

substitutes. 

t' = assigned time from the present to the maturity of one of 

the substitutes. 

When the common difference of time is to apply to all the common 
substitutes, and is to be measured from the present, 

s = S -f- n, and t = T -f- [£(n -f 1)]. 

Prop. 45. — An investment of $2,100 having 90 days to matu- 
rity is to be substituted by 3 others of like sums each, which are 
to become due at the close of a common difference of time from 
the present, and from one to another : the common sum and com- 
mon difference of the times are required. 

2100 — 3 = $700, the common sum of the substitutes ; and 

90 — 2 = 45 days, the common difference, or common interval. 

The three substitutes of $700 each, therefore, are to be made 
payable, the first at the expiration of 45 days, the second at the 
expiration of 90 days, and the third at the expiration of 135 days 
from the present time. 

Proof 
700 X 45 = 31500 
700 X 90 = 63000 
700 X 135 = 94500 
21 ) 189 ooo = 9 I or ™0 X (45+ 90 + 135) = 2100 X90. 

Prop. 46. — Five notes are to be made for like sums each, and 
are to become due at the close of equal intervals of time from the 
present, and from one to another ; and these notes are to be given 
in exchange for the four following obligations ; viz., $1600, due in 
90 days, $1250.62 due in 80 days, $852.21 due in 57 days, and 
$1865 due in 175 days, from the present time. The common de- 



INVESTMENTS, MIXED NEGOTIATIONS, ETC. 135 

nomination of the notes, and the common interval of time are 
required. 

1600.00 X 90 = 144000 

1250.62 X 80 = 100050 
852.21 X 57= 48576 

1865.00 X 175 = 326375 

5567.83 ) 619001 = 111 days, the mean or average time 

of maturity from the present of the obligations to be converted ; 
and 

5567.83 -f- 5 = $1113.57, the common denomination of the 
notes ; and 

111 -7-^(5 -}- 1) ■=. 37 days, the common difference of time, or 
common interval. 

The special times to maturity, therefore, of the five notes of 
$1113.57 each, are 37, 74, 111, 148, 185 days from the present 
time. 

When one of the common substitutes is to be treated as cash, or is to 
be considered as due at the present time, and the common interval 
is to be measured from the present, 

s = S-±-n, and t= T-±[±(n — 1)]. 
Prop. 47. — Several matters of indebtedness, amounting in 
the aggregate to $2175.44, and which will collectively mature, or 
become due by average, at the close of 68 days from the present 
time, are to be cancelled by the payment of one-fourth of their 
sum down, and by passing three notes, made for one-fourth of 
their sum each, and payable at the close of a common difference 
of time from the present and from one to another. The common 
sum and common difference are required. 

2175.44 -f- 4 = $543.86, the common sum ; and 
68 -j- ^(4 — 1) = 68 -j- 1.5 = 46 days, the common interval. 
The three notes, therefore, are to be made for $543.86 each, and 
are to be made payable, the first at 46, the second at 92, and the 
third at 138 days from the present time. 

When the common interval is to apply between all the common sub- 
stitutes, and is to be measured from an assigned time for the matu- 
rity of one of them, 

s = S^-n, and t = (T ~ t') -^- [±(n — 1)]. 
Prop. 48. — A debt of $4500, due 123 days hence, without in- 
terest, is to be substituted by 4 notes, made for one-fourth of the 
sum each, which are to run an equal interval of time from matu- 
rity to maturity, and one of them is to fie made payable at the 
close of an interval of 30 days from the present time. The com- 
mon sum and common interval are required. 
2 



145 INVESTMENTS, MIXED NEGOTIATIONS, ETC. 

4500 -f- 4 = $1125, the common denomination of the notes ; and 

(123 — 30) -f- 1.5 = 62 days, the common difference of time. 

The times to maturity of the substitutes, therefore, are 30, 92, 
154, 216 days later than the present time. 

Prop. 49. — The last preceding proposition (Prop. 47), except 
that the given debt of $4500 has but 75 days to maturity ; and 
one of the notes is not to become due until the lapse of 105 days 
from the present time. 

(75 ~ 105) -f- 1.5 = 20 days, the common interval. 

The times from the present to the maturities of the notes, there- 
fore, are 45, 65, 85, 105 days. 

Prop. 50. — It is proposed to relinquish obligations, amounting 
in the aggregate to $2554.72, and which will collectively become 
due by equation at the close of 40 days from the present time, and 
to receive in their place 3 notes, made for one-fourth of the sum 
each, and the balance in ready cash ; the said notes to run equal 
intervals of time from maturity to maturity, measured from the 
present, and one of them to run 85 days ; the adjusting interest, 
or discount, to be at 7 per cent. 

(85 — 40) -f--|(4 — 1) = 30 days, the common interval. 

The three notes of $638.68 each, therefore, are to be made pay- 
able at 85, 55, 25 days ; and the present worth of the cash pay- 
ment of $638.68, due 30 — 25 = 5 days hence, is $638.07. 

To invest a given sum of money in parts, at unlike rates of interest^ 
and the parts to gain like interest in equal intervals of time. 

Pi P' -> Pi & c - = tne parts, or partial investments. 

S = the sum of the investments. 

r, /, r", &c. ==the given rates, or these in their relations to 

each other, expressed in any proportion preferred. 

m = the product of r, r, r", &c, as expressed. 

at xi r Til . to , m e 

N== the sum of - -f- -,-+—, > &c - 

P , P ; f, to. = ^ = *» L j™, J, inversely. 

Prop. 51. — Twenty thousand dollars ($20,000) are to be 
placed at interest in three such parts that the interest on them, 
at their respective rates of 6, 7, and 8 per cent, a year, shall be 
equal for all like intervals of time. The parts, or special invest- 
ments, are required. 

to == 6 X 7 X 8 = 336 ; and 336 -f- 6 = 56 
336-^-7 = 48 
336-1-8 ==42 

"146, the value of N; . • . 



INVESTMENTS, MIXED NEGOTIATIONS, ETC. b 15 

20000 X 56 20000 X 28 
P= 146 = 73 =«7,671.28 

20000 X 24 
P'= 73 ' ' * 

20000 X 21 



73 



= 5,753.43 



$20,000.00 



To invest a given sum of money in parts, at like rates of interest, and 
for unequal intervals of time ; and the parts to gain like interest 
at the close of their respective times. 

t, t', t", &c. = the given times, or these in their relations to 
each other, expressed in whatever proportion preferred. 
m = product of t, t', t", &c, as expressed. 

i\r = sumof « + « + « &c. 
t ' if ' t" 

S, and p, p', p", &c, as in the last preceding proposition. 

, „ Sm Sm Sm . , 

p, p', p", &c, = M = m = m , &c-, inversely. 

Prop. 52. — It is proposed to place $25,000 at interest in four 
separate sums, one of them for 60, one for 80, one for 110, and one 
for 150 days' time ; and that these sums shall be such, that, at like 
rates of interest, they will gain like interest at the close of their 
respective times. The special sums are required. 

m=60 X 80 X HO X 150, or 6 X 8 X H X 15, or 
3 X 4 X 5.5 X 7.5 = 495 ; and 
495-^-3 =165. 
495 -f-4 =123.75 
. 495-^-5.5= 90. 
495-f-7.5=r 66. 

444.75 = N; therefore 
25,000X165 = 25,000 X 33 = 
r 444.75 88.95 

, _ 25,000 X 12375 _ 25,000 X 495 _ 6 956 16 
P ~4t±±lb 1779 

25,000X90 = 2 

444.75 



_ 25,000 X 66 
444.75 



= 3,709.95 

$25,000.00 



16 b INVESTMENTS, MIXED NEGOTIATIONS, ETC. 

To invest a given sum of money in parts, at unlike rates of interest, 
and for unequal intervals of time ; and the parts to gain like inter- 
est at the close of their respective times, 
m = product of the given times, or of their relations to each 

other, by any measure whatever, that is common to them ; or of 

the given rates, if preferred. 

S, JV, p, p', p", &c, as in the preceding. 

v r>> r>" &c — Sm — Sm — ^- &c 

Pi P i P •> v^- 7^7 A7V , — atTTJ/ ' 

Ntr Ntr Nt"r" 

Prop. 53. — Ten thousand dollars ($10,000) are to be placed 

at interest in three separate sums, one of them at 4 per cent., for 

240 days ; one at 6 per cent., for 120 days ; and one at 8 per cent., 

for 80 days ; and these sums are to be such that they will gain 

like interest, one with another, at the close of their respective 

times. The special sums are required. 

240X120X80, or 24X12X8, or 6X3X2 = 36 = tXt'Xt"=m. 

4, 6, 8, or 2, 3, 4, = r, r', r" 

12, 9, 8, the products of tr, tY, 

t"r", and 

36-f-12 = 3 

36-f- 9=4 

36 -^ 8 = 4.5 

11.5 = 2V, therefore 

, 10,000 X 30 _ 10,000 X 6 _ 7n 

r 115 23 

10,000 X40_ 10,000 X 8 _ 2(J 

115 23 

1^000X45 = 10 > 000X9 = ^ $10 ooom 

r 115 23 ' 

To invest a given sum of money in parts, at like rates of interest, 

and for unequal intervals of time ; and the amount (principal 

and interest) of the parts to be equal at the close of their respective 

times. 

Phop. 54. — It is proposed to place $16,000 at interest in four 

separate sums, each at 7 per cent, a year : one of them for 80, one 

for 100, one for 150, and one for 200 days' time; and that these 

sums shall be such that their amount shall be equal, one with 

another, at the close of their respective times. The special sums 

are required. 

Ar n 365 , 365 , 365 s 1 s 

iv = sum of — + ; + 5 &c, = , &c. 

365 + ^365 4- rt' ^365 -{- rt" ' l+-&s 



INVESTMENTS, MIXED NEGOTIATIONS, ETC. b 17 

' n & _ 3655 _ 3655 3655 

p, p , p , c. _ ________ _ _________ _ __________ 

&e., = , &c. : then 

1 J_ rt ' ' 

"T 1T6 "5 

365-^370.6 __ 0.98489 
365-7-372.' =-=0.98118 
365 -^375.5=_= 0.97204 
365-^379. =0.96306 



3.90117, the value of _V; and 
5J4W0 __ S4039>36 
^ 370.6iV 

5,840,000^ 402416 
372_V 
,__ 5^840^000 __ 398665 
375.5iV 

p"' = _____ = 3949.83. $16,000.00 
3792V ' 

Therefore, A = 5 -J- _V_= $4101.33, the common amount. 
If the times, £, Z', £", &c, be taken in years instead of days, then 

2V=-J_ + -_i — -f __, &c. ; and p = S -f- _V(1 -f rt) ; 

i>'= 5 -7- 2V(1 -f-rt'), & c - And, if tne times be taken in months, 

125-7- _V(1 2 + rt 7 ), &c. 

To invest a given sum of money in parts, at unlike rates of interest, 
and for unequal intervals of time ; and the amount of the parts to 
be equal, one with another, at the close of their respective times. 
Prop. 55. — The last preceding proposition, except that the 
rates are to be 6 per cent, for the 80 days' term, 7 per cent, for the 
100 days' term, 8 per cent, for the 150 days' term, and 9 per cent, 
for the 200 days' term, instead of 7 per cent, for each of the terms. 
365-^369.8 = 0.98702 
365-^-372. =0.98118 
365-^377. =0.96817 
365^-383. =0.95300 

3.88937, the value of N; and 
p = 5,840,000 -7- 369.87V = $4060.38 
p' = 5,840,000 -i- 37 22V = 4036.37 
p" = 5,840,000 -f- 37 72V = 3982.83 

3832V = 3920.44. $16,000.02. 
2* 



18 & INVESTMENTS, MIXED NEGOTIATIONS, ETC. 

Prop. 5Q. — Ganson & Co. are paying $12,000 a year in 
monthly payments in advance for the use of the premises they oc- 
cupy ; and they wish to know how much less the rent would be 
per annum, were it fixed at the same price by the year, but pay- 
able quarterly, at the expiration of each quarter-year ; allowing 
money to be worth 7 per cent, interest a year. 

P + Pr ( n + *> = 12p + pr ( n + ^ = $12,455.00, the amount, 
2n 2 

by the present manner of paying, at the close of the year ; and 

P-f- Prn = 12p+^ 12 "" w ) = $12,315.00, the amount, 
2(n-}-l) 2 

by the proposed manner of paying, at the close of the year, in 
both of which cases n represents the number of payments to be 
made per annum ; then 

12455 — 12315 = $140.00. Ans. 



SECTION I. 

MONEYS, WEIGHTS AND MEASURES, 

OF THE UNITED STATES ; — THEIR DENOMINATIONS, VALUES, 
COMPARATIVE VALUES, MAGNITUDES, Ac. 

MONEYS OP ACCOUNT OF THE UNITED STATES. 

These are the mill, the cent, the dime, and the dollar. 
10 mills =s 1 cent, 10 cents = 1 dime, 10 dimes = 1 dollar. 

The dollar is the unit or ultimate money of account of the United 
States, or of what is sometimes called Federal money. 

In practice, the dime, as a denomination of value, is rejected. 
Thus, 

10 mills = 1 cent, and 100 cents = 1 dollar. 

This mark, $, is equivalent to the word dollar, or dollars, in this 
money. 

COINS OF THE UNITED STATES. 

Until June, 1834, the government of the United States estimated 
gold in comparison with silver as 15 to 1 , and in comparison with 
copper as 850 to 1. 

From June, 1834, until February, 1853, the same government 
estimated gold in comparison with silver as 16 to 1, and in com- 
parison with copper as 720 to 1. 

For all time since February, 1853, this government has estimated 
gold in comparison with silver as 14^§f to 1, and in comparison 
with copper as 720 to 1. 

The standard for mint gold with this government until 1834. 
was 11 parts pure gold and 1 part alloy, the alloy to consist of 
silver and copper mixed, not exceeding one half copper. 

The gold coins, therefore, struck at the United States mint prior 
to 1834, are 22 carats fine. 
2 



14 CURRENCY OF THE UNITED STATES. 

In what, until 1834,' constituted a dollar of gold coin of United 
States mintage, there were put 24.75 grains of pure gold ; and 27 
grains of the standard mint gold of that day were at that time worth 
$1. Twenty-seven grains of that gold, or gold of that standard, 
are now, by the present government standard of valuation, worth 
$1.0652. 

The standard for mint silver with this government until 1834, 
was 1485 parts pure silver and 179 parts pure copper, = 8^7% 
parts pure silver and 1 part pure copper. 

The silver coins, therefore, struck at the United States mint prior 
to 1834, are lOf f f ounces fine. 

In that which, until 1834, constituted a dollar of silver coin of 
this government's mintage, there were put 371^ grains of pure 
silver; and 416 grains of the standard mint silver of that day were 
at that time of the value of $1. Four hundred and sixteen grains 
of that silver, or silver of that standard, are now, by the present 
government standard of valuation, worth $1.0744. 

The cent, until 1834, was of pure copper, and weighed 208 
grains; since 1834, pure copper, weight 168 grains. 

The standard for mint gold with this government is now, and for 
all time since June, 1834, has been, 9 parts pure gold and one part 
alloy, the alloy to consist of silver and copper mixed, not exceeding 
one half silver. 

The gold coins, therefore, struck at the United States mint and 
dated subsequent to 1834, are 21f carats fine. 

The standard weight for these coins is 25f grains to the dollar ; and 
in every 25§ grains of these coins there are 23 T 2 /^ grains of pure 
gold. 

The standard for mint silver with this government is now, and 
for all time since June, 1834, has been, 9 parts pure silver and 1 
part pure copper. 

The silver coins, therefore, struck at the United States mint and 
dated subsequent to 1834, are 10f ounces fine. 

In what, from June, 1834, until February, 1853, constituted a 
dollar of silver coin of this government's mintage, there were put 
371| grains of pure silver ; and 412£ grains of the standard mint 
silver of that day (the present standard) were worth, from June, 
1834, until February, 1853, $1. Four hundred twelve and one 
half grains of this standard of silver are now worth, by the present 
standard of valuation, $1.0742. 

The standard weight for silver coins with this government at 
present is 384 grains to the dollar. 

The foregoing is not applicable to the silver three-cent pieces, so 
called, authorized by the Congress of 1850-51. These pieces are 
composed of 3 parts silver and 1 part copper ; and their standard 
weight is 12| grains each. They are worth, at full weight, meas- 



CURRENCY OF THE UNITED STATES. 



15 



ured by the present standard of 345.6 grains of fine silver to the 
dollar, 2.685 cents each. 

The nickel and bronze tokens or coins, authorized by Congress at 
different times between the years 1856 and 1867, are as follows : — 

Nickel 1-cent token: 88 parts copper and 12 parts nickel; 
weight, 72 grains. Nickel S-cent token : copper and nickel mixed, 
not exceeding one-fourth nickel; weight, 32 grains. Nickel b-cent 
token : copper and nickel mixed, not exceeding one-fourth nickel ; 
weight, 77.16 grains, or 5 grammes; diameter, 2 centimetres. 

Bronze 1-cent token : 95 j>arts copper, and 5 parts tin and zinc ; 
weight, 48 grains. Bronze 1-cent token : 95 parts copper, and 5 
parts tin and zinc ; weight 96 grains. 

Note. — In the preceding calculations of values, and in the following, as is 
now the common custom everywhere, no value has heen assigned to the alloy, 
either in the silver coins or the gold coins. In general, it is simply copper, and 
will not net more, it is assumed, than the cost of recovering. In the United States, 
the law relating to coinage, previous to 1834, required that the alloy for gold 
coins should consist of silver and copper mixed, not exceeding one-half copper; 
and the present law provides that it shall consist of silver and copper mixed, 
not exceeding one-half silver. 
Boston, July, 1869. 

GOLD, — PURE. 
24 carats fine = pure gold. 
1 grain = $0.^3066. 

23.22 grains = SI. 00. 



dwt. 
ounce 



= $1.033592. 

= $20.671834. 



MINT GOLD, — U. S. 



Alloy, practically all copper. 
Nine parts of pure gold and one part of alloy, or 
Ine = standard coin. 
= $0.03876. 
= $1.00. 
= $0.93023. 
— $18.60465. 



1 grain 

25-| grains 

1 dwt. 

1 ounce 



GOLD COINS, — U. S. 



Denominations. 


Weight in 
grains. 


Standard 
value. 


Double Eagle, 


_ 


516 


$20.00 


Eagle, - 




258 


10.00 


Half-Eagle, - 


_ 


129 


5.00 


Quarter-Eagle, - 


_ 


v -64^ 


2.50 


Triple Gold Dollar, 


_ 


77| 


3.00 


Gold Dollar, 


_ 


25| 


1.00 


Eagle, prior to 1834 


(com. value = $10.62), 


270 


10.909 


Half do. " « 


(com. value = $5.31), 


135 


5.454 



16 



CURRENCY OF THE UNITED STATES. 



Private and Uncurrent. 






Weight in 
Grains. 


Sales. 


A. Bechtler, N. C, $5 piece, 




$4.75 


" 2£ " - 


- 


- 




2.37 


(< << 1 (( - 


. 


. 




.93 


T. Reed, Georgia, 5 " - 


- 


- 




4.75 


« " 2i " - 


- 


- 




2.37 


a n 1 " - 


. 


- 




.93 


Moffat, California, 5 " - 


- 


- 


129 


5.00 



12 ounces fine = Pure Silver. 
1 dwt. = $0.06944. 

345.6 grains = $1. 
1 ounce =5 | 



MINT SILVER. — U. S. 

Alloy, all copper. 

Nine parts pure silver and one part alloy ; or 

10 oz. 16 dwts. fine = Standard Coin. 

1 dwt. * = $0.0625. 

384 grains = $1.00. 

1 ounce = $1.25. 



SILVER COINS. — U. S. 



Dollar, ------- 

Half Dollar, 

• Quarter Dollar, - - - - - 

Dime, _-___- 

Half Dime, ------ 

Three-Cent Piece, f silver and * copper, 

The copper coins of the United States are the cent and half cent ; 
they are of pure copper. The weight of the former is 168 grains, 
and that of the latter, 84 grains. 

Note. — The silver coins of the United States, issued since February, 1853, are not 
legal tender in the United States in sums exceeding,/!*;*! dollars. 



Weight in 
Grains. 


Standard 
Value. 


384 

192 

96 


$1.00 
.50 
.25 


38f 


.10 


19* 

12f 


.05 
.03 



CURRENCY OF THE UNITED STATES. 



17 



TABLE, 



Exhibiting the standard weight and present par value of the silver coins 
of the United States, of dates subsequent to 1834, and prior to 1853. 





Weight in 


Present 




Grains. 


par value. 


Dollar, - 


4124- 


$1.0742 


Half Dollar, - 


206| 


.5371 


Quarter Dollar, - - - - 


103£ 


.2685 


Dime, - 


41± 


.1074 


Half Dime, - 


m 


.0537 


Three-Cent Piece, - 


12f 


.03 



CURRENCIES OF THE DIFFERENT STATES OF THE UNION. 

1 Farthings = 1 Penny, 12 Pence = 1 Shilling, 20 Shillings =* 
1 Pound. 

In Massachusetts, Connecticut, Rhode Island, New Hampshire, 
Vermont, Maine, Kentucky, Indiana, Illinois, Missouri, Virginia, 
Tennessee, Mississippi, Texas and Florida, 6 shillings = 1 dollar ; 
$1 - A £. 

In New York, Ohio and Michigan, 8 shillings = 1 dollar ; $1 = 

In New Jersey, Pennsylvania, Delaware and Maryland, 7 shil- 
lings and 6 pence = 1 dollar ; 1 dollar = f- £. 

In North Carolina, 10 shillings s= 1 dollar ; $1 = £ £. 

In South Carolina and Georgia, 4 shillings and 8 pence = 1 dol- 
lar; $1 = ^ £. 

Note. — These currencies, so called, are nominal at present in a great measure. The 
denominations serve in the different States as verbal expressions of value. But they are 
neither the names of the moneys of account in any of the States, nor are they the national 
names of any of the real moneys in circulation. All values in money in the United State* 
are legally expressed in dollars, cents, and mills. 
2* 



18 METRICAL SYSTEM OF WEIGHTS AND MEASURES. 



THE METRICAL SYSTEM OF WEIGHTS AND MEAS- 
URES. 

In this system, the Metre is the basis, and is one forty-millionth 
of the polar circumference of the earth. 

The Metre is the principal unit measure of length ; the Are 
of surface; the Stere of solidity ; the Litre of capacity; and the 
Gram of weight. 

The gram is the weight, in a vacuum, of one cubic centimetre 
of pure water at its maximum density. 

The Metre, almost exactly . = 39.37 U. S. inches. 

The Are (100 square metres) = 3.95367 " square rods. 

The Stere (a cubic metre) . = 35.31445 « cubjc feet. 

r™ TU , ,. , .. . v (61.023 " " inches. 

The Litre (a cubic decimetre) = j 1Q5668 u w quarts< 

The Gram . = 15.43235 " grains. 

The divisions by 10, 100, 1,000, of each of these units, are ex- 
pressed by the same prefixes, viz., deci, centi, milli; and the multi- 
ples by 10, 100, 1,000, 10,000, of each, by deca, hecto, kilo, myria. 
The former series were derived from the Latin language, the latter 
from the Greek. 

To illustrate with the metre : — 

10 millimetres — 1 centimetre, 10 centimetres z= 1 decimetre, 10 
decimetres — 1 Metre, 10 Metres = 1 decametre, 10 decame- 
tres =z 1 hectometre, 10 hectometres = 1 kilometre, 10 kilometres 
— 1 myriametre. 

In commerce, the ordinary weight is the kilogram, and 100 kilo- 
grams (usually called kilos) = 1 quintal; 10 quintals z=z 1 millier, 
or tonneau. The kilogram = 15,432.35 -^ 7000 ±= 2.20462 avoir- 
dupois pounds. 

In practice, the terms milliare, declare, decare, kiliare, and myri- 
are are usually dropped, and 

100 centare = 1 are ; T.00 ares = 1 hectare. 

Also the terms minister e, hectostere, kilistere, and myriastere, are 
usually rejected, and 1 00 centisteres =: 1 decistere; 10 decisteres 
=: 1 stere ; 10 steres = 1 decastere == 353.1445 cubic feet. 
1 centiare (square metre) = 1.19598526 square yards. 
1 kilometre . . . = 0.62137 statute miles. 
1 hectare . . . = 2.471 = U. S. acres. 

1 kilolitre . . = 1 stere = 61,023.377953 cubic in. 

A hectolitre = 26.41748 wine gallons = 2.83774 Winchester bush. 

Note. — The system is the one recommended by the Statistical Congress of 
1865 as a general system of weights and measures to be adopted by all nations. 



IOREIGN GOLD COINS. 



19 



FOREIGN GOLD COINS. 



Note. — The coins of any country, both, gold and silver, circulating as for- 
eign in any other, particularly those of the smaller denominations, are usually 
held at an estimate below their standard par value, compared with the money 
standard of the country in which they circulate as foreign. Many of them, 
more particularly the silver, having circulation in the United States, are much 
worn and otherwise depreciated. In some instances, owing to frequent changes 
made both with regard to weight and purity, certain of them, having the same 
name and general appearance, bear a premium at home; others, a discount. 
Others, again, can hardly be said to have a definable value anywhere. The par 
value of the old pistole of Geneva, for instance, weighing 103i grains, is $3,985, 
while that of the new, weighing 87$ grains, would, at the same degree of 
purity, be worth but $3,386 ; whereas, owing to its higher standard of fine- 
ness, its par value is $3,443. The ducat of Austria, coined in 1831, weighs 
531 grains, — its purity is 23.64, and its par value $2,269; while the half 
sovereign, closely resembling the ducat, coined in 1835, and weighing 87 grains, 
has a purity only of 21.64, and a par value, consequently, of but $3,378. The 
circulating value of the ducat in the United States, in general, is $2.20, and 
that of the half sovereign of Austria, $3.25. 





Standard 


Standard 


Par value 


Circulating- 


Par val- 


ARGENTINE REPUBLIC. 


of 
purity in 


weight 
in 


in 
Federal 


value in 
Federal 


ue per 


Doubloon to 1832, 


carats. 


grains. 

418 


money. 


money. 


cU. 


19.56 


$14,671 


$ 


3.50 


" to " 


20.83 


415 


15.512 




3.73 


AUSTRIA. 












Sovereign, half in propor- 












tion, to 1785, 


22.00 


170 


6.711 


6.50 


3.94 


Sovereign, half in propor- 












tion, since 1785, 


21.64 


174 


6.756 


6.50 


3.88 


Ducat, double in propor- 












tion, 


23.64 


53i 


2.269 


2.20 


4.24 


BELGIUM. 












Sovereign, half in pro., 


22.00 


170 


6.711 




3.94 



20 



FOREIGN GOLD COINS. 





Standard 


Standard 


Par value 


Circulating 


Par v il- 


• 


flf 


weight 


in 


value in 


ue per 




purity in 


in 


Federal 


Federal 


grain. 


Twenty Franc, more in pro. 


carats. 

2L50 


grains. 


money. 


money. 

$3.83 


els. 


99£ 


$3,840 


3.85 


Ducat, 








2.20 




Bolivia, Colombia, Chili, 












Ecuador, Peru, New- 












Grenada, and Mexico. 












For the modern coins, 












&c, of these States, see 












Foreign Moneys of Ac- 












count, Sec. A. 


* 










Doubloon, (8 E) 


20.86 


417 


15.620 


15.60 


3.74 


Half do. 


ic 


208£ 


7.810 


7.50 


c« 


Quarter do. 


u 


1044 


3.905 


3.75 


U 


Eighth do. 


tt 


52 


1.952 


1.75 


Ct 


Sixteenth do. 


it 


26 


.976 


.90 


cc 


Pistole, half in pro., 








3.75 




BRAZIL. 












For the modern coinage 












of this Empire, see For- 












eign Moneys of Account 












and Coins, Sec. A. 












Dobraon, 


22.00 


828 


32.719 


32.00 


3.95 


Dobra, 


a 


438 


17.306 


17.00 


. a 


Joannes, {standard variable) 


a 


432 


17.064 


$13to$17 


u 


Half do. do. do. 


<< 


216 


8.532 


$6 to 8.50 


tt 


Moidore, (BBBB) half in 












pro., {standard variable) 


21.79 


165 


6.451 


6.00 


3.90 


Crusado, do. do. 


a 


16| 


.635 




it 


DENMARK. 












Christian d'or 


21.74 


103 


4.018 




3.90 


Ducat, species, 


23.48 


53£ 


2.254 


2.20 


4.21 


" current, 


21.03 


48 


1.811 




3.77 


FRANCE. 












There are but few gold 












coins of France now in 












circulation other than 












multiples of the standard 












franc, napoleons, frac- 












tional and double, in pro. 












Chr. d'or, double in pro., 


21.60 


101 


3.914 


3.90 


3.87 



FOREIGN GOLD COINS. 



21 



Franc d'or, double in pro., 
Louis d'or, " " " 

to 1786, 
Louis d'or, double in pro., 

since 1786, 
Napol©Dn(20F.) double &c. 


Standard 

of 

purity in 

cirats. 


Standard 
weight 

in 
grains. 


Par value 

Federal 
money. 


Circulating- 
value in 
Federal 
money. 


Par val- 
ue per 
grain. 


21.60 

21.49 

21.68 
21.60 


101 
125£ 

118 

99| 


$3,914 

4.840 

4.573 

3.856 


$3.90 

4.50 
3.83 


3.87 
3.85 
3.87 


GERMANY. 












BADEN. 

Zehn Gulden, 5 in pro., 


21.60 


105£ 


4.088 


4.00 


3.87 


BAVARIA. 

Carolin, 

Ducat, double in pro., 

Maximilian, 


18.49 
23.58 
18.49 


149i 
53| 
100 


4.952 
2.275 
3.317 


2.20 


3.32 
4.23 
3.31 


BRUNSWICK. 

Ducat, 

Pistole, double in pro., 

Ten Thaler, 5 in pro., to 

1813, 
Ten Thaler, less in pro., 

since 1813, 


23.22 
21.60 

21.55 

21.50 


53i 
117i 

202 

204 


2.220 
4.548 

7.811 

7.873 


7.80 
7.80 


4.16 

3.87 

3.86 
3.85 


HANOVER. 

Ducat, double in pro., 
George d'or, " " " 
Zehn Thaler, 5 " " 


23.83 

21.67 
21.36 


53£ 
102h 
204£ 


2.287 
3.987 
7.838 


2.20 
7.80 


4.27 

3.88 
3.83 


HESSE. 

Ten Thaler, 5 in pro., to 

1785, 
Ten Thaler, 5 in pro., since 

1785, 


21.36 
21.41 


202 
203 


7.742 
7.799 




3.84 


SAXONY. 

Ducat, 

Augustus d'or, double in 
pro., since 1784. 


23.49 
21.55 


53£ 
102£ 


2.256 
3.964 


2.20 


4.21 
3.86 


WURTEMBURG. 

Carolin, 
Ducat, 


18.51 
23.28 


147£ 
53£ 


4.899 
2.235 




3.32 
4.17 



22 



FOREIGN GOLD COINS. 





Standard 


Standard 


Par value 


Circulating 


Par val- 




of 


weight 


in 


value in 


ue per 




purity in 


in 


Federal 


Federal 


grain. 


GREAT BRITAIN. 


carats. 


grains. 


money. 


money. 


cts. 












(Alloy, since 1826, all copper.) 












The modern gold coins of 












this Kingdom are the 












sovereign, fractional, 












double, &c. 












Guinea, half in pro., to 












1785, 


22.00 


127 


$5,016 




3.95 


Guinea, half in pro., since 












1785, 


a 


129£ 


5.111 


$5.00 


(t 


Sovereign, half in pro., 


n 


123£ 


4.866 


4.83 


a 


Five do. 


C( 


616^ 


24.332 


24.20 


a 


Sovereign, (dragon) half 












in pro., 


tt 


122* 


4.838 


4.80 


a 


Double Sovereign (dragon) 


(« 


246 


9.717 


9.67 


a 


GREECE. 












Twenty Drachm, more in 












pro., 


21.60 


89 


3.441 


3.40 


3.87 


HOLLAND. 












Ducat, 


23.58 


53£ 


2.263 


2.20 


4.23 


Ryder, 


22.00 


153 


6.043 




3.95 


Double do. 


a 


309 


12.205 




it 


Ten Gulden, 5 in pro., 


21.60 


103 


3.988 


3.98 


3.87 


INDIA. 












Pagoda, star, 


19.00 


52| 


1.798 




3.40 


Mohur, (E. I. do.) 1835. 


22.00 


180 


7.106 


6.75 


3.95 


Half Sovereign, do. 








2.41 




BOMBAY. 












Rupee, 


22.09 


179 


7.095 




3.96 


MADRAS. 












Rupee, 


22.00 


180 


7.106 




3.95 


ITALY. 












Eturia, Ruspone, 


23.97 


16U 


6.935 




4.30 


Genoa, Sequin, 


23.86 


53£ 


2.291 




4.28 


Milan, Pistole, 


21.76 


97£ 


3.807 




3.90 


" Sequin, 


23.76 


53k 


2.281 




4.26 















FOREIGN GOLD COINS. 



23 





Standard 


Standard 


Par value 


Circulating 


Par val- 




of 


weight 


in 


value in 


ue per 




purity in 


in 


Federal 


Federal 


grain. 


Milan, Twenty Lire, more 


carats. 


grains. 


money. 


money. 


els. 












in proportion, 


21.58 


99£ 


$3,853 


$3.83 


3.86 


Naples, Ducat, multiples 












in pro., 


21.43 


22£ 


.865 




3.84 


Naples, Oncetta, 


23.88 


58 


2.485 




4.28 


Parma, Doppia, to 1786, 


21.24 


110 


4.192 




3.81 


" Pistole, since 1796, 


20.95 


110 


4.135 




3.75 


" Twenty Lire, 


21.60 


99£ 


3.859 


3.83 


3.87 


Piedmont, Carlino, half in 












pro., since 1785, 


21.69 


702 


27.321 




3.89 


Piedmont, Pistole, half in 












pro., since 1785, 


21.54 


140 


5.411 




3.86 


Piedmont, Sequin, half in 












pro., since 1785, 


23.64 


53| 


2.280 




4.23 


Piedmont, Twenty Lire, 












more in pro., 


20.00 


99| 


3.563 


3.50 


3.59 


Rome, Ten Scudi, 5 in pro. 


21.60 


267£ 


10.368 




3.87 


" Sequin, since 1760, 


23.90 


52k 


2.251 




4.28 


Sardinia, Carlino, £ in 












pro., 


21.31 


247£ 


9.465 




3.82 


Tuscany, Zechino, double 












in pro., 


23.86 


531 


2.302 




4.30 


Venice, Zechino, double 












in pro., 


23.84 


54 


2.310 






MALTA. 












Sequin, 


23.70 


53£ 


2.275 




4.25 


Louis d'or, double and 












demi in pro., 


20.25 


128 


4.651 




3.63 


NETHERLANDS. 












Ducat, 


23.52 


53£ 


2.257 




4.21 


Zehn Gulden, 5 in pro., 


21.55 


1031 


4.013 


4.00 


3.86 


POLAND. 












Ducat, 


23.58 


53£ 


2.264 




4.23 


PORTUGAL. 












The modern Portuguese 












gold coins are the coroa 












of 5000 reis, parts and 












multiples in proportion. 












See Sec. A. 











24 



FOREIGN GOLD COINS. 





Standard 


Standard 


Par value 


Circulating 


Par val- 




of 


weight 


in 


value in 


ue per 




purity in 


in 


Federal 


Federal 


grain. 


Dobraon, 24,000 reis, 


carats. 


grains. 


money. 


$32.00 


cts. 


22.00 


828 


$32,706 


3.95 


Dobra, 


it 


438 


17.301 


17.00 


it 


Joannes, {standard variable) 


it 


432 


17.064 


$13 to $17 


it 


Half " " " 


it 


216 


8.532 


$6 to 8.50 


it 


Moidore, 4000 reis, " 








84* to $4} 




Coroa, 5000 " 


cc 


147i 


5.83 


5.75 


a 


Milrea, 


22.00 


19| 


.780 




3.95 


PRUSSIA. 












Ducat, 


23.49 


53£ 


2.255 


2.20 


4.21 


Frederick d'or, double in 












pro., 


21.60 


102£ 


3.973 




3.87 


RUSSIA. 












Ducat, 


23.64 


54 


2.291 




4.24 


Imperial, (10 R.) half in 












pro., 1801, 


23.55 


185| 


7.828 




4.22 


Imperial, (10 R.) half in 












pro., since 1818, 


22.00 


201£ 


7.949 


7.90 


3.95 


SICILY. 












Oncia, double in pro., 


20.39 


68i 


2.495 




3.64 


Twenty Lire, more in pro., 


21.60 


99£ 


3.856 


3.83 


3.87 


SPAIN. 












For the new standard of 












coinage, denominations, 












&c, of this Kingdom, 












see Foreign Moneys of 












Account and Coins, Sec. 
A. 
Doubloon (8 S) parts in pro. 












21.45 


416£ 


16.031 


16.00 


3.84 


" (8 E) parts as 












Bolivian, &c.^ 


20.86 


417 


15.620 


15.60 


3.74 


Pistole, to 1782, 


21.48 


103 


3.970 




3.85 


" since " 


20.93 


104 


3.906 




3.75 


Escudo, to 1788, 


20.98 


52 


1.957 




3.76 


" since " 


20.42 


52 


1.905 




3.66 


Coronilla " 1800, 


20.29 


27 


.983 




3.64 


SWEDEN. 












Ducat 


23.45 


53 


2.230 




4.20 



LONG OR LINEAR MEASULE. 



25 



SWITZERLAND. 
Berne, Ducat, double in 

pro., 
Berne, Pistole, 
Geneva, Pistole, 

" " (old) 

Zurich, Ducat, double in 
pro., 



Standard 


Standard 


Par value 


Circulating 


of 


weight 


in 


value in 


purity in 


in 


Federal 


Federal 


carats. 


grains. 


money. 


money. 


23.53 


47 


$1,984 




21.62 


117£ 


4.558 




21.87 


87a 


3.443 




21.51 


103i 


3.985 




23.50 


53£ 


2.256 




15.88 


36* 


1.040 




19.25 


53 


1.830 




22.88 


73| 


3.027 





Par val 

ue per 



4.22 
3.88 
3.92 
3.85 

4.21 



TURKEY. 

Misseir, half in pro. 1820, 15.88 36£ 1.040 2.84 

Sequin fonducli, 19.25 53 1.830 3.45 

Yeermeeblekblek, 22.88 73| 3.027 4.10 

Note. — For full and particular specifications regarding modern foreign 
gold and silver coins, see Foreign Moneys of Account and coins, Sec. A. 

The standard silver 5-franc piece of France is worth $1.00471 in the silver 
coins of the United States ; hut 5 francs in the standard gold coins of France 
are worth hut .$0.964726 in the gold coins of the United States. 



LONG OR LINEAR MEASURE. — U. S. 

Standard. — A brass rod, the length of which, at 62° Fahrenheit, 
is -|f ;y§§-§ that of a pendulum beating seconds in vacuo, at the level 
of the sea, at the latitude of London, = f|;^ao at 32° Fah., at the 
gravitation at New York, = the Yard. 



6 points 

12 lines (72 points) 

13 inches 

3 feet (36 inches) 



= 1 line. 
= 1 inch. 
= 1 foot. 
= 1 yard. 



5h yards (16^ ft.) =1 rod. 
40 rods (220 yds.) = 1 furlong. 
8 fur. (5280 feet) = 1 stat. mile. 



SPECIAL, FOR CLOTH. 
2| inches = 1 nail. 

4 nails (9 inches) = 1 quarter. 



4 quarters (36 inches) = 1 yard. 



7 T 9 ^r inches 
25 links 



SPECIAL, FOR LAND. 

= 1 link. I 100 links (66 feet) = 1 chain. 

— 1 rod. I 80 chains (320 rods) = 1 s. mile. 

engineer's chain. 

10 inches = 1 link. 

120 links (100 feet) = 1 chain. 



26 



SQUARE OR SUPERFICIAL MEASURE. 



SHOEMAKER'S MEASURE. 
No. 1 is 4J inches in length, and each succeeding number is an 
addition of £ of an inch. No. 1 man's size = 8£i inches. 



MISCELLANEOUS. 



Hair's breadth 
Digit 
Palm 
Hand 



= ^ 1 F inch. 
= 10 lines. 
= 3 inches. 
=== 4 " 
9 



Fathom = 6 feet. 

Knot = 47| feet. 

Cable's length = 120 fathoms. 
Geometrical pace= 4.4 feet. 



12 particular things = 1 dozen. 
12 dozen (144) = 1 gross. 

12 gross (1728) = 1 great gross. 
20 particular things = 1 score. 
24 sheets of paper = 1 quire. 
20 quires = 1 ream. 

SQUARE OR SUPERFICIAL MEASURE. 
{Length X breadth.) 
144 square inches = 1 square foot. 
9 " feet =1 " yard. 
30| " yards = 1 " rod. 
40 " rods = 1 rood. 

4 roods = 1 acre. 

SPECIAL, FOR LAND. 
62§|4 square inches = 1 square link. 



10000 
10 

Square rod 

Rood 



links = 1 
chains = 1 



chain. 

acre. 

272^ square feet. 



1210 

10890 " 

4840 " 

43560 " 

640 
102400 sq. 
220 X 198 square feet ~| 

The square of 12.649 " rods 

" " of 69.5701 " 
" " of 208.710321 " feet 



Acre ( 160 square rods) = 
Square mile = 



yards. 

feet. 

yards. 

feet. 

acres. 

rods. 



= 1 acre. 



CUBIC OR SOLID MEASURE. 



27 



CIRCULAR MEASURE. 



Minute, or " 
Geogra- 
phical m. 
(60") 

League 

Degree 



I 1.152 s. miles. 
— \ 6086 feet. 

= 3 miles. 
_ { 60 geo. miles. 
~~ ') 69.158 s. ms. 



24897 s. m. 



Great Circle = 360 degn 
Equatorial cir- 
cumference 
of the earth 
Equatorial diam.= 7925 
Polar diam. = 7899 
Mean radius = 3955.92 



H 



Sign( T ^ zod.)=30 degrees. 

Note. — In the expressions, square feet and feet square, there is this difference ; viz., 
the former expresses an area in which there are as many square feet as the number 
named, and the latter an area in which there are as many square feet as the square of the 
number named. The former particularizes no form of area, the latter asserts a square 
form. 

CUBIC OR SOLID MEASURE. — U. S. 



Cubic foot, 
1728 cu. inches 



Cylindrical foot 
1728 " inches 



(Length X breadth X depth.) 

f 1.273 cylindrical feet. 

J 2200 " inches. 

) 3300 spherical " 

l_ 6600 conical " 

f 0.785398 cubic feet. 

j 1357.2 " inches. 
] 2592 spherical " 
l_5184 conical 
1 cubic yard. 



h 



27 cubic feet = 



of round timber = 1 ton. 



" of shipping 
" of hewn 



40 
42 

50 
128 " 

Cubic foot of pure water, "i 
at the maximum density 
at the level of the sea, 
(39°.83, barometer 30 
inches) 

Cylindrical foot 

Cubic inch 



= 1 ton. 
= 1 ton. 
= 1 cord. 



Cylindrical inch 

Pound 

" distilled 

Cubic inch " = 

Pound at 62°, distilled = 

Cubic inch at 62°, " = 

" 39°.83, in vacuo = 



= { 

_ J 49.1 

~~ } 785.- 

-! 

- i 



62£ avoirdupois pounds. 



ounces. 



0.036169" 

0.5787 " 

253.1829 
028415 avd. 
4546 
27.648 cubic inches. 
27.7015 « " 
252.6934 grains. 
27.7274 cub. inches. 
252.458 grains. 
253.0864 " 



pounds. 

ounces. 

pounds. 

ounces. 

grains. 

pounds. 

ounces. 



Cubic foot of salt water (sea) weighs 64.3 pounds. 



28 



GENERAL MEASURE OF WEIGHT. 



GENERAL MEASURE 

AVOIRDUPOIS. 

Standard. — The pound is the 
weight, taken in air, of 27.7015 
cubic inches of distilled water at 
its maximum density, (39°. 83 F., 
the barometer being at 30 inches) 
= 27.7274 cubic inches of distilled 
water at 62° = 7000 Troy grains. 

27-g-^- grains = 1 dram. 

16 drams (437£ grs.) = 1 ounce. 
16 ounces (7000 grs.)= 1 pound. 



SPECIAL 

28 pounds = 

4 quarters > 
112 pounds 5 
20 cwt. = 



GROSS. 



1 quarter. 
5 1 quintal. 



cwt. 
1 ton. 



SPECIAL ' 



DIAMOND. 



16 parts = 1 grain : 
4 grs. = 1 carat 



0.8 troygr. 
3.2 " " 



OF WEIGHT. — U. S. 

SPECIAL TROY. 

(Exclusively for gold and sil- 
ver bullion, precious stones, and 
gold, silver and copper coins, and 
with reference to their monetary 
value only.) 

24 grains = 1 pennyw't. 

20dwts. (480 grs.)= 1 ounce. 
12 oz. (5760 grs.) = 1 pound. 

SPECIAL APOTHECARIES'. 

(Exclusively for compounding 
medicines, for recipes and pre- 
scriptions.) 
20 grains = 1 scruple, 9. 

3 scruples = 1 dram, 3. 

8 drams(480 g.)= 1 ounce, B. 
12 oz. (5760 g.) = 1 pound, lb, 

1 lb. avoir. = ly 3 ^ lbs. troy. 
1 lb. troy = \$% lbs. avoir. 
1 oz. avoir. = -f|-|- oz. troy. 



1 oz. troy 



1 TT 7 5 oz. avoir. 



Note. — The comparative value of diamonds of the same quality is as the square of 
their respective weights. A diamond of fair quality, weighing 1 carat in the rough state, 
is estimated worth about $9^^ ; and it will require one of twice that weight to make 
one when worked down equal to 1 carat in weight. Hence, to determine the value of a 
wrought diamond of any given number of carats : — Rule. — Double the weight in carats 
and multiply the square by 9.50. Thus, the value of a wrought diamond, weighing 2 
carats, is 2 + 2=4 X 4 = 16 X9.50=K152. ' 



LIQUID MEASURE. — U. S. 

The " Wine" or " Winchester" Gallon, of 231 cubic inches 
capacity, is the Government or Customs gallon of the United States 
for all liquids, and the legal gallon of each state in which no law 
exists fixing a state or statute gallon of its own. It contains 58372£ 
grains of distilled water at 39°. 83, the barometer being at 30 inches. 

4 gills = 1 pint, 2 pints = 1 quart. 
4 quarts, or 231 cubic in. ) _ 5 1 gallon. 

\ ™ \ 8.355 av'd. lbs. pure water. 



0.13368 cub. ft., 294.1176 cyl. in. 



DRY MEASURE. 29 



■ 0.128 cubic foot. 
Liquid gallon of the } 1221.184" in. 



State of New York, --,- ~ g avoid ^ re water 
281.62 cylmdnc in. ) . o Q0 «, u £ n in _ 



Barrel = 31£ gallons. 
Tierce = 42 
Hogshead = 63 " 



Puncheon = 84 gallons. 
PipeorBatt = 126 " 
Tun = 252 " 



H 



211.211 cub. in. $ "" \ at 62° F. , b. 30 in. 

Ale gallon, ) _ ( 10^ av'd lbs. pure water 
282 cub. in. > ) at 39°. 83, b. 30 in. 



(0.8331 

)n = { 0.8191 

(0.1014 



0.8331 Imperial gallon. 
Wine gallon = { 0.8191 Ale " 

^42 W. bushel. 



1 Imperial gallon = 1.2 Wine gallons. 



DRY MEASURE. — U. S. 

The " Winchester Bushel," so called, of 2150 T ^. cubic inches 
capacity, is the Government bushel of the United States, and the legal 
bushel of each state having no special or statute bushel of its own. 
The standard Winchester bushel measure is a cylindrical vessel hav- 
ing an outside diameter of 19£ inches, an inside diameter of 18£ 
inches, and an inside depth of 8 inches. The standard " heaped " or 
" coal" bushel of England was this measure heaped to a true cone 6 
inches high, the base being 19£ inches, or equal to the outside diam- 
eter of the measure. Its ratio to the even bushel was, therefore, as 
1.28, nearly, to 1. The present " Imperial " measure of England has 
the same outside diameter and the same depth as the Winchester, and 
an internal diameter of 18.8 inches, and the same height of cone is 
retained for forming the heaped bushel. Its ratio, therefore, to the 
even bushel is a trifle less than was that of the Winchester. In the 
United States the " heaped bushel " is usually estimated at 5 even 
pecks, or as 1.25 to 1 of the standard even bushel, which, if taken as 

* By enactment of the Legislature of the State of New York, this gallon ceased to be the 
legal gallon of that State, April 11, 1852 ; and the United States Government gallon, of 231 
cubic inches capacity, was adopted in its stead. 

3# 



30 



DRY MEASURE. 



the rule, requires a cone on the Winchester measure of 5.4 inches to 
equal the heaped Winchester bushel. 



4 gills . 
2 pints . 

4 quarts 

8 quarts 
4 pecks 

2150.42 cubic in. 
1.244456 " ft. 
1.5844 cyl. " J 

Bushel of the ) 

State of New York ,*} 
2816.1955 cyl. in. ) 

Bushel of Connecticut,! 



— \ 7 



Heaped Win. bushel 
1.28— even" " 
Imperial bushel 
Chaldron 

1 Winchester bushel 

1 Imperial bushel 



-{ 

i- 



1 pint. 

1 quart. 
K 1 gallon 
\ or half peck. 

I peck. 

1 bushel. 

2738 cyl. in. 
, 77.7785 av'd lbs. 
Lpure water. 

f 1.28 cubic feet. 
[ 2211.84 " in. 

80 av'd lbs. pure water. 

1.272 cubic feet. 

2198 " in. 

79.50 av'd lbs. pure water. 
S 2747.7 cubic in. 
\ 1.59 cubic ft. 

2218.192 " in. 

36 Winch, heaped bushels. 
0.9694 Imperial bushel. 
9.3092 Wine gallons. 

1.0315 Winchester bushels. 



Note. — The Imperial bushel, mentioned above, is the present legal bushel of Great 
Britain ; and the Imperial gallon, mentioned on the preceding page, is the present legal 
gallon of Great Britain, for all liquids. The gallon for liquids is the same as the gallon 
for dry measure. Eight Imperial gallons make one bushel. The subdivisions of the gal- 
lon and the bushel, and their denominations, are the same as in the Winchester measures. 
In Great Britain, in addition to the denominations of dry measure used in the United 
States, the 



Strike, = 2 bushels. 

Coomb, = 4 " 

Quarter, = 8 " 

Wey or load, = 40 " 



Last, = 80 bushels. 

Sack of corn, = 3 " 

Bole of corn, = 6 " 

Last of gunpowder, . . = 42 barrels. 



* This bushel ceased to be the legal bushel of this State April 11, 1852, and the United 
States Government bushel, of 2150-j^y cubic inches oapacity, was adopted as the legal 
bushel in its stead. 

t This bushel is now, January, 1852, no longer the legal bushel of this State, and the 
standard Winchester bushel is adopted in its stead. 



SECTION II. 

MISCELLANEOUS FACTS, CALCULATIONS, AND PRACTICAL 
MATHEMATICAL DATA. 



SPECIFIC GRAVITIES. 
The specific gravity of a body is its weight relative to the weight 
of an equal bulk of pure water at the maximum density, (39°. 83, b. 
30 in.) the water being taken as 1., a cubic foot of which weighs 
1000 avoirdupois ounces, or 62£ lbs. The specific gravity, therefore, 
of any body multiplied by 1000, or, which is the same thing, the dec- 
imal being carried to three places of figures, or thousands, as in the 
following tables, the whole taken as an integer equals the number 
of ounces in a cubic foot of the material : multiplied by 62.5, or con- 
sidered an integer and divided by 16, it equals the number of pounds 
in a cubic foot; and multiplied by .036169, or taken as an integer 
and divided by 27648, it equals the decimal fraction of a pound per 
cubic inch ; by which, it is readily seen, the specific gravity of a 
commodity being known, its weight per any given bulk is easily and 
accurately ascertained ; as, also, its specific gravity, the weight and 
bulk being known. The weight of any one article relative to that of 
any other, is as its respective specific gravity to the specific gravity 
of the other. 



METALS. 


Specific 
gravity. 




Specific 
gravity. 


Antimony, . 


6.712 


Gold, pure, hammerec 


19.546 


Arsenic, 


5.810 


Iridium, 


15.363 


Bismuth, 


9.823 


Iron, cast, 


7.209 


Bronze, 


8.700 


" wrought, . 


7.787 


Brass, best, . 


8.504 


Lead, 


11.352 


Copper, cast, 


8.788 


Mercury, 32°, . 


13.598 


" wire-drawn, 


8.878 


" ' 60°, . 


13.580 


Cadmium, . 


8.604 


" —39°, . 


15.000 


Cobalt, 


7.700 


Manganese, 


8.013 


Chromium, 


5.900 


Molybdenum, 


8.611 


Glucinium, 


3.000 


Nickel, . 


8.280 


Gold, pure, cast, . 


19.258 


Osmium, . . , 


10.000 



32 



SPECIFIC GRAVITIES. 





Specific 




Specifio 




gravity. 




gravity. 


Platinum, cast, . 


19.500 


Granite, red, 


2.625 


" hammered, 


20.337 


' ' Lockport, 


^.655 


rolled, 


22.069 


" Quincy, 


2.652 


Potassium, 60°, . 


0.865 


" Susquehanna, 


2.704 


Palladium, 


11.870 


Grindstone, . 


2.143 


Rhodium, 


11.000 


Gypsum, opaque, 


. • 2.168 


Silver, pure, cast, 


10.474 


Hone, white, 


2.876 


' ' hammered, 


10.511 


Hornblende, 


3.600 


Sodium, 


0.970 


Ivory, 


1.822 


Steel, soft, 


7.836 


Jasper, 


2.690 


" tempered, 


7.818 


Limestone, green, 


3.180 


Tin, cast, 


7.291 


" white, 


3.156 


Tellurium, 


6.115 


Lime, compact, . 


2.720 


Tungsten, 


17.600 


" foliated, . 


2.837 


Titanium, , 


4.200 


" quick, 


0.804 


Uranium, 


9.000 


Loadstone, . 


4.930 


Zinc, cast, 


6.861 


Magnesia, hyd., . 


2.333 






Marble, common, 


2.686 


STONES AND EA 


RTHS. 


" white Ital. 


2.708 


Alabaster, white, 


2.730 


" Rutland, Vt., 


2.708 


" yellow, 


2.699 


" Parian, . 


2.838 


Amber, 


1.078 


Nitre, crude, 


1.900 


Asbestos, starry, 


3.073 


Pearl, oriental, . 


2.650 


Borax, 


1.714 


Peat, hard, 


1.329 


Bone, ox, . 


1.656 


Porcelain, China, 


2.385 


Brick, 


1.900 


Porphyra, red, 


2.766 


Chalk, white, 


2.782 


" green, 


2.675 


Charcoal, . 


.441 


Quartz, 


2.647 


" triturated, 


1.380 


Rock Crystal, 


2.654 


Cinnabar, . 


7.786 


Ruby, 


4.283 


Clay, 


1.934 


Stone, common, . 


2.520 


Coal, bitum. avg., 


1.270 


" paving, 


2.416 


" anth. " 


1.520 


" pumice, 


0.915 


Coral, red, 


2.700 


" rotten, 


1.981 


Earth, loose, 


1.500 


Salt, common, solid, 


2.130 


Emery, 


4.000 


Saltpetre, refined, 


2.090 


Feldspar, 


2.500 


Sand, dry, . 


. 1.800 


Flint, white, 


2.594 


Serpentine, 


2.430 


" black, 


2.582 


Shale, . . . 


2.600 


Garnet, 


4.085 


Slate, 


2.672 


Glass, flint, 


2.933 [ Spar, fluor, 


3.156 


" white, 


2.892 Stalactite, . 


2.324 


" plate, 


2.710 


Tale, black, 


2.900 


M green, 


2.642 


Topaz, 


4.011 



SPECIFIC GRAVITIES. 



33 





Specific 
gravity. 




Specific 
gravity. 


9 SIMPLE SUBSTANCES, 


Pine, yellow, 


.568 


neither metallic 


nor gaseous. 


Poplar, white, 


.383 


Boron, 


1.968 


Plum, . 


.785 


Biomine, 


2.970 


Quince, 


.705 


Carbon, 


3.521 


Spruce, white, 


.551 


Iodine, 


4.943 


Sassafras, 


.482 


Phosphorus, 


1.770 


Sycamore, 


.604 


Selenium, . 


4.320 


Walnut, 


.671 


Silicon, 


1.184 


Willow, 


.585 


Sulphur, 


1.990 


Yew, Spanish, 


.807 






" Dutch, 


.788 


WOODS, 


(dry.) 






Apple, 


0.793 


Highly seasoned Am. 


Alder, 


.800 


Ash, white, . 


. . .722 


Ash, . 


.760 


Beech, 


.624 


Beech, 


.696 


Birch, 


.526 


Birch, 


.720 


Cedar, . 


.452 


Box, French, 


1.328 


Cherry, 


.606 


" Dutch, 


.912 


Cypress, 


.441 


Cedar, 


.561 


Elm, . 


.600 


Cherry, 


.715 


Fir, . 


.491 


Chestnut, 


.610 


Hickory, red, 


.838 


Cocoa, 


1.040 


Maple, hard, 


.560 


Cork, 


.240 


Oak, white, uplanc 


I, . .687 


Cypress, 


.644 


" James River, 


.759 


Ebony, American 


1.331 


Pine, yellow, 


.541 


" foreign, 


1.290 


" pitch, . 


.536 


Elm, . 


.671 


" white, 


.473 


Fir, yellow, 


.657 


Poplar, (tulip,) 


.587 


" white, 


.569 


Spruce, white, 


.465 


Hacmetac, . 


.592 






Hickory, red, 


.900 


GUMS, FAT 


3, &C. 


Lignum vitae, 
Larch, 


1.333 
.544 


Asphaltum, . 


5 .905 
• I 1.650 


Logwood, . 


.913 


Beeswax, 


.965 


Mahogany, Spanis 


sh, best, 1.065 


Butter, 


.942 


a a 


com., .800 


Camphor, 


.988 


St. Dc 


uningo, .720 


Gamboge, 


1.222 


Maple, red, 


.750 


Gunpowder, . 


.900 


Mulberry, . 


.897 


" shake] 


i, . 1.000 


Oak, live, . 


1.120 


" solid, 


< 1.550 
* \ 1.800 


" white, 


, .785 


Orange, 


.705 


Gum, Arabic, 


. 1.454 


Pear, 


.661 


" Caoutchouc. 


.933 


Pine, white, 


.554 


" Mastic, 


. 1.074 



34 



SPECIFIC GRAVITIES. 



Honey, 

Ice, . 

Indigo, 

Lard, 

Pitch, 

Rosin, 

Spermaceti, 

Starch, 

Sugar, dry, 

Tallow, 

Tar, . 

LIQUIDS. 

Acid, acetic, 

" citric, 

" fluoric, 

" nitric, 

11 nitrous, 

" sulphuric, 

" muriatic, 

" silicic, 
Alcohol, anhyd. 
90 % 
Beer, 

Blood, human, 
Camphene, pure, 
Cider, whole, 
Ether, sulph., 

" nitric, 
Milk, cow's, 
Molasses, 75 % 
Oils, linseed, 

" olive, 

" rapeseed, 

" sassafras, 

" turpentine, com. 

" sperm, pure, 

" whale, p'f'd, 
Proof spirits, 
Vinegar, 
Water, pure, 

" sea, 

'* Dead sea,, 



Specific 
gravity. 

1.450 

.930 
1.009 

.941 
1.253 
1.100 

.943 
1.530 
1.606 

.938 
1.257 



1.062 

1.034 

1.060 

1.485 

1.420 

1.846 

1.200 

2.660 

.794 

.834 

1.034 

1.054 

.863 

1.018 

.715 

.908 

1.032 

1.400 

.934 

.917 

.927 

1.090 

.875 

.874 

.923 

.925 

1.025 

1.000 

1.026 

1.240 



Wine, champagne, 
" claret, 
" port, 
" sherry, 



Specific 

gravity. 

.997 
.994 
.997 
.992 



ELASTIC FLUIDS: 

The measure of which is atmospheric air, 
at 60°, b. 30 in., its assumed gravity 1 ; one 
cubic foot of which weighs 527.04 grains, = 
.305 of a grain per cubic inch. It is, at 
this temperature and density, to pure water 
at the maximum density, as .0012046 to 1, 
or as 1 to 830.1. 



SIMPLE OR ELEMENTARY GASES. 



Hydrogen, 


.0689 


Oxygen, . 


1.1025 


Nitrogen, . 


.9720 


Fluorine, . 




Chlorine, 


2.470 


Carbon, vapor of, 


I .422 


(theoretically,) 


COMPOUND GAS 


JES. 


Ammoniacal, 


.591 


Carbonic acid, . 


1.525 


" oxide, . 


.973 


Carbureted hydrogen, 


.559 


Chi oro-carbonic, 


3.389 


Cyanogen, . 


1.818 


Muriatic acid gas, 


1.247 


Nitrous acid gas, 


3.176 


Nitrous oxide gas, 


1.040 


Olefiant, gas 


.978 


Phosphureted hydroge 


1, 1.185 


Sulphureted " 


1.177 


Steam, 212° 


.484 


Smoke, of wood, 


.900 


" of coal, 


.102 


Vapor, of water, 


.623 


" of alcohol, 


1.613 


" of spirits turpei 


itine, 5.013 



WEIGHT PER BUSHEL BARREL GALLON, &C. 35 

Weight per Bushel (corn Ex. tariff) of different Grains, Seeds, 6fc. 



Articles. 


Ihs. 


Articles. 




Ihs. 


Barley, (N. E. 47 lbs.) . 48 


Hemp seed, 


. 


Beans, ■ . 


62 


Oats, 




32 


Buckwheat, 


46 


Peas, 




64 


Blue-grass seed 


14 Rye, 




56 


Corn, 


56 Salt, T. L, 




80 


Cranberries, 


| " boiled, . 




56 


Clover seed, 


60 Timothy seed, 




46 


Dried Apples, 


22 Wheat, . 




60 


" Peaches, 


33 Potatoes, h'p'd, 




60 


Flax seed, (N. E. 52 lbs.) 56 i Malt, ". * .' 




38 


Weight per Barrel (Legal or by Usage) of different Articles. 


Flour, . . . 196 lbs. 


Cider, in Mass., 


32 gals. 


Boiled 3alt, . . 280 " 


Soap, 


256 lbs. 


Beef, . . . 200 " 


Raisins, . 


112 " 


Pork, ... 200 " 


Anchovies, 


30 " 


Pickled Fish, . . 200 " 


Lime, 


220 " 


" in I 


Ground Plaster, 




Massachusetts, 5 

A Gallon of Oil wei^ 


OKJ glH. 


Hydraulic Cement, . 


300 " 


?hs 


7| lbs. 


A " " Molasses, standard, (75 per cent.,) 


HI " 


A " " Linseed Oil (usage, 1\ lbs.) . 7.788 " 


A Firkin of Butter, (legal,) 


56 " 


A Keg of powder, 


25 « 


A Hogshead of Salt is 


8 bush. 


A Perch of Stone = 24| cubic feet. 




A Gallon of Alcohol, 90 per cent., weighs . 


6.965 lbs. 


A " " Proof Spirits, " 


7.732 » 


A " " Wine, (average,) " 


8.3 


A " " Sperm Oil, " 


7.33 " 


A " " Whale " p'f 'd, " 


7.71 " 


A " " Olive " " 


7.66 » 


A " " Spirits Turpentine, " 


7.31 » 


A " " Campher 


le, pure 


> • 


7.21 


<( 



Weight of Coals, fyc, broken to the medium size, per Measure of 
Capacity. 

The average weight of Bituminous Coals, broken as above, is about 
62 per cent, that of a bulk of equal dimensions in the solid mass, or 



36 



ROPES AND CABLES. 



of the specific gravity of the article ; that of Anthracite is about 
5 7 per cent. 



Average weight ) 
per oubio foot. J 


lbs. 


Average weight per \ 
W. Coal bushel, j 


Anthracite, . 


54 


Anthracite, 


Bituminous, . 


50 


Bituminous, 


Charcoal, of pine, . 


18.6 


Charcoal, hard wood 


" of hard wood, . 


19.02 


Coke, best, 



lbs. 

86 
80 
30 
32 



Practical Approximate Weight in Pounds of Various Articles, 

95 



Sand, dry, per cubic foot, 
Clay, compact, per cubic foot, 
Granite, " " " 

Lime, quick, " " " 

Marble, " " " 

Slate, " " " 

Peat, hard, " " " 

Seasoned Beech Wood, per cord, 

" Yellow Birch Wood, per cord, . 

" Red Maple Wood, " " . 

" " Oak Wood, 

" White Pine Wood, 

" Hickory Wood, 

" Chestnut Wood, 
Meadow Hay, well settled, per cubic foot, 81 lbs., or 
240 cubic feet z= 2000 lbs., or 26 8 T 8 ^ cubic feet 
zzz 1 long ton 
Meadow Hay, in large old stacks, per cubic foot, 
Clover Hay, in settled bulk, " " " 

Corn on Cob, in crib, " " " 

" shelled, in bin, " " " 

Wheat, in bin, " " " 

Oats, in bin, " " " 

Potatoes, in bin, " " " 

Common Brick, 7| X 3f X H in. " M, 
Front " 8X4|X2| in. " " 



135 

165 

50 

169 

167 

83 

5616 

4736 

5040 

6200 

4264 

6960 

4880 



25* 



4500 
6185 



ROPES AND CABLES. 

The strength of cords depends somewhat upon the fineness of 
the strands ; — damp cordage is stronger than dry, and untarred 
stonger than tarred ; but the latter is impervious to water and less 
elastic. 

Silk cords have three times the strength of those of flax of 
equal circumference, and Manilla has about half that of hemp. 



WEIGHT AND STRENGTH OF IRON CHAINS. 



3T 



Ropes made of iron wire are full three times stronger than those 
of hemp of equal circumference. 

White ropes are found to be most durable. The best qualities of 
hemp are — 1. pearl gray; 2. greenish; 3. yellow. A brown color 
has less strength. 

The breaking weight of a good hemp rope is 6400 lbs. per square 
inch, but no cordage may be counted on with safety as capable of sus- 
taining a weight or strain above half that required to break it, and 
the weight of the rope itself should be included in the estimate. 

The reliable strength of a good hemp cable, in pounds, is usually 
estimated as equal to the square of its circumference in inches X by 
120. That of rope X 200. Thus, a cable of 9 inches in circumfer- 
ence may be relied on as having a sustaining power = 9 X 9 X 120 
= 9720 lbs. 

The weight, in pounds, of a cable laid rope, per linear foot = the 
square of its circumference in inches X .036, very nearly. 

The weight, in pounds, of a linear foot of manilla rope = the 
square of its circumference in inches X -03, very nearly. Thus, a 
manilla rope of three inches circumference weighs per linear foot 
3 X 3 X -03 = T 2 ^ lbs., = 3^ feet per lb. 

A good hemp rope stretches about %, and its diameter is diminished 
about •-§• before breaking. 

WEIGHT AND STRENGTH OE IRON CHAINS. 



Diameter of 


Weight of 


Breaking 


Diameter of 


Weight of 


Breaking 


Wire 


Hoot 


Weight 


, Wire 


lFoot 


Weight 


in Inches. 


of Chain. 


of Chain. 


in Inches. 


of Chain. 


of Chain. 




lbs. 


lbs. 




lbs. 


lbs. 


A 


0.325 


2240 


t 


4.217 


26880 


i 


0.65 


4256 


« 


4.833 


32704 


A 


0.967 


6720 


* 


5.75 


38752 


3 

1" 


1.383 


9634 


if 


6.667 


45696 


A 


1.767 


13216 


I 


7.5 


51744 


* 


2.633 


17248 


it 


9.333 


58464 1 


i_* 


3.333 


21728 


l 


10.817 


65632 j 



38 



COMPARATIVE WEIGHT OF METALS. 



Comparative Weight of Metals, Weight per Measure of Solidity, %c 





Specific 


Ratio of 


Pounds 


n a Cubic 


Iron, wrought or rolled 


Gravity. 


Comparison 


Foot. 


Inch. 


7.787 


1. 


486.65 


.28163 


Cast Iron, 


7.209 


.9258 


450.55 


.26073 


Steel, soft, rolled, . 


7.836 


1.0064 


489.75 


.28342 


Copper, pure, " 


8.878 


1.1401 


554.83 


.32110 


Brass, best, " 


8.604 


1.1050 


537.75 


.3112 


Bronze, gun metal, . 


8.700 


1.1173 


543.75 


.31464 


Lead, .... 


11.352 


1.4579 


709.50 


.4106 



TABLE, 

Exhibiting the Weight in pounds of One Foot in Length of Wrought 
or Rolled Iron of any size, (cross section,) from £ inch to 12 inches. 



SQUARE BAR. 



Size 


Weight 


Size 


Weight 


Size 


Weight 


Size 


Weight 


Inches. 
J 


Pounds. 


Inches. 

2f 


Pounds. 


Inches. 


Pounds. 


Inches. 


Pounds. 


.053 


19.066 


4| 


72.305 


71 


203.024 


i 


.211 


24 


21.120 


41 


76.264 


8 


216.336 


1 


.475 


2£ 


23.292 


45 


80.333 


8| 


230.068 


h 


.845 


2| 


25.560 


5 


84.480 


84 


244.220 


£ 


1.320 


25 


27.939 


5J 


88.784 


81 


258.800 


i 


1.901 


3 


30.416 


5* 


93.168 


9 


273.792 


I 


2.588 


3i 


33.010 


5f 


97.657 


9* 


289.220 




3.380 


3i 


35.704 


5* 


102.240 


94 


305.056 


U 


4.278 


3f 


38.503 


5* 


106.953 


91 


321.332 


n 


5.280 


34 


41.408 


5| 


111.756 


10 


337.920 


n 


6.390 


3f 


44.418 


5* 


116.671 


10* 


355.136 


u 


7.604 


31 


47.534 


6 


121.664 


104 


372.672 


ii 


8.926 


3* 


50.756 


H 


132.040 


lot 


390.628 


ii 


10.352 


4 


54.084 


64 


142.816 


n 


408.960 


14 


11.883 


4* 


57.517 


6| 


154.012 


ii* 


427.812 


2 


13.520 


H 


61.055 


7 


165.632 


114 


447.024 


21 


15.263 


41 


64.700 


7k 


177.672 


HI 


466.684 


n 


17.112 


4£ 


68.448 


n 


190.136 


12 


486.656 



COMPARATIVE WEIGHT OF METALS. 



39 



To determine the weight, in pounds, of one foot in length, or of any 
length, of a bar of any of the following metals of form prescribed, of 
any size, multiply the weight in pounds, of an equal length of square 
rolled iron of the same size, (see table of square rolled iron,) if the 
weight be sought of 

Iron, Round rolled, by 7854 

Steel, Square " " 1.0064 

Round " " 7904 

Cast Iron, Square bar, " . . . . ■, .9258 

" " Round " " 7271 

Copper, Square rolled, " 1.1401 

Round " " 8954 

Brass, Square " " ..... 1.105 

" Round " " 8679 

Bronze, Square bar, " 1.1173 

" Round " " 8775 

Lead, Square " " 1.4579 

Round " " 1.145 

The weight of a bar of any metal, or other substance, of any given 
length, of a flat form, (and any other form maybe included in the 
rule,) is readily obtained by multiplying its cubic contents (feet or 
inches) by the weight (pounds, ounces, or grains) of a cubic foot or 
inch of the article sought to be weighed ; that is — 

Length X breadth X thickness X weight per unit of measure. 

For the weight in pounds of a cubic foot or inch of different metals, 
see " Table of weights of metals per measure of solidity, &c." 

OR, FOR FLAT OR SQUARE BARS, 

Multiply the sectional area in inches by the length in feet, and that 
product, if the metal be 

Wrought Iron, by 3.3795 

Cast " "...... 3.1287 

Steel, " 3.4 

Example. — Required the weight of a bar of steel, whose length is 
7 feet, breadth 2£ inches, and thickness | of an inch. 

2.5 X .75 X 7 X 3.4 = 44.625 lbs. Ans. 

Example. — Required the weight of a cast iron beam, whose length 
IB 14 feet, breadth 9 inches, and thickness 1£ inch. 

14 X 9 X 1-5 X 3.1287 = 591.32 lbs. Ans. 



40 



WEIGHT OF ROUND ROLLED IRON. 



TABLE, 

Exhibiting the weight in pounds of One Foot in Length of Round Rolled 
Iron of any diameter , from \ inch to 12 inches. 



Diameter 


Weight 


Diam. in 


Weight 


Diam. in 


Weight 


Diam. in 


Weight 


in inches. 


in lbs. 


inches. 


in lbs. 


incheS. 


in lbs. 


inches. 


in lbs. 

159.456 


1 


.041 


2| 


14.975 


4| 


56.788 


71 


tk 


.165 


2£ 


16.688 


4| 


59.900 


8 


169.856 


ft 


.373 


2| 


18.293 


4& 


63.094 


H 


180.696 


h 


.663 


2| 


20.076 


5 


66.752 


H 


191.808 


I 


1.043 


2& 


21.944 


5i 


69.731 


81 


203.260 


I 


1.493 


3 


23.888 


54 


73.172 


9 


215.040 


i 


2.032 


H 


25.926 


6| 


76.700 


n 


227.152 




2.654 


H 


28.040 


^ 


80.304 


9£ 


239.600 


n 


3.360 


8| 


30.240 


5| 


84.001 


91 


252.376 


u 


4.172 


3£ 


32.512 


5| 


87.776 


10 


266.288 


if 


5.019 


3f 


34.886 


5& 


91.634 


104 


278.924 


u 


5.972 


31 


37.332 


6 


95.552 


10* 


292.688 


it 


7.010 


8| 


39.864 


64 


103.704 


101 


306.800 


it 


8.128 


4 


42.464 


6£ 


112.160 


11 


321.216 


H 


9.333 


4J 


45.174 


61 


120.960 


114 


336.004 


2 


10.616 


. 44 


47.952 


7 


130.048 


Hi 


351.104 


2J 


11.988 


4f 


50.815 


n 


139.544 


HI 


366.536 


24 


13.440 


J 4i 


53.760 


n 


149.328 


12 


382.208 



To find the weight of an equilateral three-sided cast iron prism. 
width of side in inches X 1> 354 X length in feet = weight in lbs. 

Example. — A three-sided east iron prism is 14 feet in length, and 
the width of each side is 6 inches ; required the weight of the prism. 

6 2 X 1-354 X 14 = 682.4 lbs. Arts. 
To find the weight of an equilateral rectangular cast iron prism. 

width of side in inches" X 3.128 X length in feet = weight in lbs. 
To find the weight of an equilateral five-sided cast iron prism. 

width of side in inches X 5-381 X length in feet = weight in lbs. 
To find the weight of an equilateral six-sided cast iron prism. 

width of side in inches 2 X 8.128 X length in feet = weight in lbs. 
To find the weight of an equilateral eight-sided cast iron prism 

width of side in inches X 15.1 X length in feet = weight in lbs. 
To find the weight of a cast iron cylinder. 

diameter in inches" X 2.457 X length in feet = weight in lbs. 

In a quantity of cast iron weighing 125 lbs., how many cubio 
inches ? 

By tabular weight per cubic inch — 

125 -i- .26073 =s 479.4 cubic inches. Ans % 



RELATING TO CAST IRON. 41 

Or, by tabular weight per cubic foot — 

450.55 : 1728 : : 125 : 479.4 cubic inches. Ans. 
How many cubic inches of copper will weigh as much as 479.4 
cubic inches of cast iron 1 

By tabular weight per cubic inch — 

.3211 : .26073 : : 479.4 : 389.27 cubic inches. Ans. 
Or, by specific gravities — 

8.878 : 7.209 : : 479.4 : 389.27 cubic inches. Ans. 
Or, by tabular ratio of weight — 
.9258 
479.4 X HioT = 389 - 28 * 
A cast ircm rectangular weight is to be constructed having a 
breadth of 4 inches and a thickness of 2 inches, and its weight is 
to be 18 lbs. ; what must be its length ? 
18 
4X2X.26073 =ss8 * 63inches - Ans ' 
A cast iron cylinder is to be 2 inches in diameter, and is to weigh 
6 lbs. ; what must be its length ? 

.26073 X .7854 =.2047 lb. = weight of 1 cyl. inch, then 
6 
2 2 X- 2047 = 7 ' 327 mches - Ans - 
A cast iron cylinder is to weigh 6 lbs., and its length is to bo 
7.327 inches ; what must be its diameter 1 

*^( 7.327 X -2047 ) = 2 inches. Ans. 

A cast iron weight, in the form of a prismoid, or the frustrum of 
a pyramid, or the frustrum of a cone, is to be constructed that will 
weigh 14 lbs., and the area of one of the bases is to be 16 inches, 
and that of the other 4 inches ; what must be the length of the 
weight 1 

14 



Y* 1 * 6 X 4 - 8 and 8 + 16 + 4 + 3 - 9.33, and - 33 mn 
= 5.75 mches. Ans. ^ 

Note. — For Rules in detail pertaining to the foregoing, see Geometry, Mensuration 
of superficies — of solids. 

A model for a piece of casting, made of dry white pine, weighs 
7 lbs. ; what will the casting weigh, if made of common brass 1 
By specific gravities — 

.554 : 8.604 : : 7 : 108.71 lbs. Ans. 

Note. — As the specific gravity of the substance of which the model is composed must 
generally remain to some extent uncertain, calculations of this kind can only be relied on 
as approximate. 

4* 



TABLE 

Exhibiting the Weight of One Foot in Length of Flat, Rolled Iron ; 
Breadth and Thickness in Inches, Weight in Pounds. 



Br. and Th. 


Wei't. 


Br. and Th. 


Wei't. 


Br. and Th. 


Wei't. 


Br. and Th. 


Wei't. 


inch. 


lbs. 


inch. 


lbs. 


inch. 


lbs. 


inch. 


lbs. 


h by 4 


.211 


U by I 


3.696 


1| by I 


2.957 


24 by ft 


3.591 


i 


.422 


l 


4.224 


1 


3.696 


f 


4.488 


f 


.634 


li 


4.752 


a 

4 


4.435 


1 


5.386 


f by J 


.264 


If by J 


.581 


i 


5.175 


1 


6.284 


4 


.528 


i 

4 


1.161 




5.914 


1 


7.181 


% 


.792 


f 


1.742 


n 


6.653 


14 


8.079 


h 


1.056 


h 


2.323 


i i. 

J -4 


7.393 


14 


8.977 


1 by \ 


.316 


f 


2.904 




8.132 


if 


9.874 


i 

4 


.633 


1 


3.485 


n 


8.871 


H 


10.772 


I 


.950 


I 


4.066 


if 


9.610 


24 by 4 


.950 


h 


1.267 


1 


4.647 


11 by ^ 


.792 


4 


1.901 


f 


1.584 


lg 


5.228 


4 


1.584 


f 


2.851 


I by 4 


.369 


14 


5.808 


1 


2.376 


h 


3.802 


i 

4 


.739 


U by 4 


.634 


1 


3.168 


f 


4.752 


f 


1.108 


4 


1.267 


f 


3.960 


1 


5.703 


h 


1.478 


1 


1.901 


1 


4.752 


I 


6.653 


f 


1.848 


k 


2.534 


I 


5.544 


1 


7.604 


1 


2.218 


I 


3.168 


1 


6.336 


14 


8.554 


1 by | 


.422 


1 


3.802 


1| 


7.129 


14 


9.505 


4 


.845 


1 


4.435 


u 


7.921 


11 


10.455 


§ 


1.267 


1 


5.069 


If 


8.713 


li 


11.406 


£ 


1.690 


n 


5.703 


u 


9.505 


If 


12.356 


1 


2.112 


n 


6.337 


If 


10.297 


It 


13.307 


1 


2.534 


it 


6.970 


11 


11.089 


2f by 4 


1.003 


§ 


2.957 


If by J 


.686 


2 by 4 


.845 


4 


2.006 


1J by 4 


.475 


i 

4 


1.373 


4 


1.690 


f 


3.010 


4 


.950 


I 


2.059 


f 


2.534 


h 


4.013 


f 


1.425 


£ 


2.746 


£ 


3.379 


f 


5.016 


£ 


1.901 


f 


3.432 


f 


4.224 


I 


6.019 


1 


2.376 


a 

4 


4.119 


1 


5.069 


§ 


7.023 


1 


2.851 


i 


4.805 


1 


5.914 


1 


8.026 


I 


3.326 


1 


5.492 


1 


6.759 


14 


9.029 


1 


3.802 


H 


6.178 


n 


7.604 


14 


10.032 


Id by 4 


.528 


H 


6.864 


14 


8.449 


If 


11.0301 


4 


1.056 


if 


7.551 


if 


9.294 


14 


12.030] 


I 


1.584 


U 


8.237 


u 


10.138 


If 


13.042 


h 


2.112 


1| by 4 


.739 


'M by i 


.898 


11 


14.040 


f 


2.640 


4 


1.478 


4 


1.795 | 


2 


10.052 


1 


3.168 


1' 


2.218 


f 


2.693 1 


2* by 4 


1.050 



WEIGHT OF FLAT, ROLLED IRON. 

TABLE. — Continued. 



43 



Br.andTh 


Weight. 


Br. and Th 


Weight. 


Br. and Th. 


Weight. 


Br. and Th. 


Weight. 


inch. 


lbs. 


inch. 


lbs. 


inch. 


lbs. 


inch. 


lbs. 

20.594 


2k by 4 


2.112 


21 by If 


16.264 


34 by | 


6.865 


3| by 1| 


f 


3.168 


11 


17.426 


1 


8.238 


H 


22.178 


h 


4.224 


2 


18.587 


I 


9.610 


14 


23.762 


I 


5.280 


2f 


19.749 


1 


10.983 


2 


25.347 


% 


6.336 


24 


20.911 


U 


12.356 


24 


28.515 


4 


7.393 


2& by J 


1.214 


14 


13.729 


2i 


31.683 




8.449 


4 


2.429 


if 


15.102 


2| 


34.851 


11 


9.505 


f 


3.644 


li 


16.475 


4 by ^ 


1.690 


H 


10.561 


i 


4.858 


If 


17.848 


4 


3.379 


if 


11.617 


f 


6.073 


11 


19.221 


i 


6.759 


li 


12.673 


1 


7.287 


14 


20.594 


i 


10.139 


if 


13.729 


I 


8.502 


2 


21.967 


l 


13.518 


li 


14.785 


1 


9.716 


24 


24.713 


14 


16.898 


1* 


15.841 


a 


10.931 


2k 


27.459 


li 


20.277 


2 


16.898 


14 


12.145 


H by J 


1.478 


H 


23.657 


2| by i 


1.109 


11 


13.360 


4 


2.957 


2 


27.036 




2.218 


li 


14.574 


f 


4.436 


24 


30.416 


I 


3.327 


11 


15.789 


i 


5.914 


2i 


33.795 


i 


4.436 


11 


17.003 


f 


7.393 


2| 


37.175 


f 


5.545 


l| 


18.218 


1 


8.871 


3 


40.555 


I 


6.653 


2 


19.432 


4 


10.350 


34 


43.934 


4 


7.762 


2£ 


20.647 


l 


11.828 


44 by i 


1.795 


1 


8.871 


24 


21.861 


n 


13.307 


4 


3.591 


14 


9.980 


3 by £ 


1.267 


14 


14.785 


i 


7.181 


n 


11.089 


4 


2.535 


if 


16.264 


a 

4 


10.772 


it 


12.198 


| 


3.802 


ii 


17.748 


1 


14.363 


ia 


13.307 


i 


5.069 


it 


19.221 


14 


17.954 


if 


14.416 


f 


6.337 


li 


20.700 


li 


21.544 


ii 


15.525 


| 


7.604 


14 


22.178 


11 


25.135 


ii 


16.634 


4 


8.871 


2 


23.657 


2 


28.726 


2 


17.742 


1 


10.139 


24 


26.614 


24 


32.317 


2£ 


18.851 


1* 


11.406 


2i 


29.571 


2i 


35.908 


2|by I 


1.162 


14 


12.673 


2| 


32.528 


2| 


39.498 


i 

4~ 


2.323 


If 


13.941 


3| by J 


1.584 


3 


13.089 


I 


3.485 


li 


15.208 


i 


3.168 


34 


16.680 


h 


4.647 


1| 


16.475 


f 


4.752 


3i 


50.271 


f 


5.808 


1| 


17.743 


i 


6.337 


4iby i 


3.802 


1 


6.970 


14 


19.010 


f 


7.921 


k 


7.604 


I 


8.132 


24 


20.277 


1 


9.505 


1 


11.406 


1 


9.294 


2 


22.812 


4 


11.089 


1 


15.208 


n 


10.455 


2i 


25.345 


1 


12.673 


14 


19.010 


14 


11.617 


34 by ^ 


1.373 


n 


14.257 


li 


22.812 


if 


12.779 


4 


2.746 


14 


15.842 


HI 


26.614 


li 


13.940 


f 


4.119 


if 


17.426 


2 


30.416 


1 if 


15.102 


i 


5.492 


li 


19.010 


24! 


34.218 



44 



WEIGHT OF FLAT, ROLLED IRON. 

TABLE. — Continued. 



Br. and Th. 


Weight. 


Br. and Th. 


Weight. 


Br. and Th. 


Weight. 


Br. and Th. 


Weight. 


inch. 


lbs. 


inch. 


lbs. 


inch. 


lbs. 


inch. 


»s. 


4* by 2k 


38.020 


4| by 3 


48.158 


54 by | 


13.307 


5k by 2 


37.175 


2| 


41.822 


34 


52.172 


1 


17.743 


2J 


46.469 


3 


45.624 


H 


56.185 


1122.178 


3 


55.762 


34 


49.426 


5 by 4 


4.224 


1£I26.614 


51 by 4 


4.858 


3£ 


53.228 


h 


8.449 


H 


31.049 


k 


9.716 


4| by | 


4.013 


1 


12.673 


2 


35.485 


1 


14.574 


•i 


8.026 


1 


16.898 


24 


39.921 


1 


19.432 


I 


12.040 


14 


21.122 


2k 


44.356 


14 


24.290 


1 


16.053 


U 


25.347 


3 


53.228 


U 


29.146 


14 


20.066 


H 


29.571 


5£ by i 


4.647 


H 


34.007 


1£ 


24.079 


2 


33.795 


h 


9.294 


2 


38.865 


l| 


28.092 


24 


38.020 


1 


13.941 


24 


43.723 


2 


32.106 


2k 


42.244 


1 


18.587 


2£ 


48.581 


24 


36.119 


3 


46.469 


14 


23.234 


3 


58.297 


2£ 


40.132 


54 by | 


4.436 


U 


27.881 


6 by 4 


5.069 


2| 


44.145 


£ 


8.871 


11 


32.528 







WEIGHT OF METALS IN PLATE. 



The weight of a square foot one inch thick of 



Malleable Iron 
Com. plate " 
Cast Iron 
Copper, wrought 

" com. plate 
Brass, plate, com. 
Zinc, cast, pure 

" sheet 
Lead, cast 



= 40.554 lbs. 

= 37.761 " 

= 37.546 " 

= 46.240 " 

= 45.312 " 

= 42.812 " 

= 35.734 " 

= 37.448 " 

= 59.125 " 



And for any other thickness, greater or less, it is the same in pro- 
portion ; thus, a square foot of sheet copper -Jg of an inch thick 
= 46.24 -f- 16 = 2.89 lbs. And 5 square feet at that thickness 
= 2.89 X 5 = 14.45 lbs., &c. So, too, 5 square feet at 2i inches 
thickness = 46.24 X 2.5 X 5 = 578 lbs. 



AMERICAN WIRE GAUGE. 



45 



THE AMERICAN WIRE GAUGE. 

The American Wire Gauge was prepared by Messrs. Brown and 
Sharp, manufacturers of machinists' tools, Providence, R. I. It is 
graded upon geometrical principles, is rapidly becoming the stand- 
ard gauge with manufacturers of wire and plate in the United 
States, and cannot fail to supersede the use of the Birmingham 
Gauge in this country. 



TABLE 



Showing the Linear Measures represented by Nos. American Wire 

Gauge and Birmingham Wire Gauge, or the values of 

the Nos. in the United-States Standard Inch. 





American 


Birm. 




American 


Birm. 




American 


Birm. 


American 


Birm. 


No. 


Gauge. 


Gauge . 


No. 


Gauge. 


Gauge 


No. 


Gauge. 


Gauge. 


No. 


Gauge. 


Gauge. 




Inch. 


Inch. 




Inch. 


Inch. 




Inch. 


Inch. 




Inch. 


Inch. 


0000 


.46000 


.454 


8 


.12849 


.165 


19 


.03589 


.042 


30 


.01003 


.012 


000 


.40964 


.425 


9 


.11443 


.148 


20 


.03196 


.035 


31 


.00893 


.010 


00 


.36480 


.380 


10 


.10189 


.134 


21 


.02846 


.032 


32 


.00795 


.009 





.32486 


.340 


11 


.09074 


.120 


22 


.02535 


.028 


33 


.00708 


.008 


1 


.28930 


.300 


12 


.08081 


.109 


23 


.02257 


.025 


34 


.00630 


.007 


2 


.25763 


.284 


13 


.07196 


.095 


24 


.02010 


.022 


35 


.00561 


.005 


3 


.22942 


.259 


14 


.06408 


.083 


25 


.01790 


.020 


36 


.00500 


.004 


4 


.20431 


.238 


15 


.05707 


.072 


26 


.01594 


.018 


37 


.00445 




5 


.18194 


.220 


16 


.05082 


.065 


27 


.01419 


.016 


38 


.00396 




6 


.16202 


.203 


17 


.04526 


.058 


28 


.01264 


.014 


39 


.00353 




7 


.14428 


.180 


18 


.04030 


.049 


29 


.01126 


.013 


40 


.00314 





Thus the diameter or size of No. 4 wire, American gauge, is 
0.20431 of an inch; Birmingham gauge, 0.238 of an inch: so the 
thickness of No. 4 plate, American gauge, is 0.20431 of an 
inch; Birmingham gauge, 0.238 of an inch; and so lor the other 
Nos. on the gauges respectively. 



TABLE 

Showing the Number of Linear Feet in One Pound, Avoirdupois, of 
Different Kinds of Wire : Sizes or Diameters corre- 
sponding to Nos. American Wire-gauge. 



No. 


Iron. 


Copper. 


Brass. 


No. 


Iron. 


Copper. 


Brass. 


Feet. 


Feet. 


Feet. 


Feet. 


Feet. 


Feet. 


0000 


1.7834 


1.5616 


1.6552 


19 


293.00 


256.57 


271.94 


000 


2.2488 


1.9692 


2.0872 


20 


396.41 


347.12 


367.92 


00 


2.8356 


2.4830 


2.6318 


21 


465.83 


407.91 


432.35 





3.5757 


3.1311 


3.3187 


22 


587.35 


514.32 


545.13 


1 


4.5088 


3.9482 


4.1847 


23 


740.74 


648.63 


687.50 


2 


5.6854 


4.9785 


5.2768 


24 


934.03 


817.89 


866.90 


3 


7.1695 


6.2780 


6.6542 


25 


1177.7 


1031.3 


1093.0 


4 


9.0403 


7.9162 


8.3906 


26 


1485.0 


1300.4 


1378.3 


5 


11.400 


9.9825 


10.581 


27- 


1872.7 


1639.8 


1738.1 


6 


14.375 


12.588 


13.342 


28 


2361.4 


2067.8 


2191.7 


7 


18.127 


15.873 


16.824 


29 


2977.9 


2607.6 


2763.8 


8 


22.857 


20.015 


21.214 


30 


3754.8 


3287.9 


3484.9 


9 


28.819 


25.235 


26.748 


31 


4734.2 


4145.5 


4394.0 


10 


36.348 


31.828 


33.735 


32 


5970.6 


5221.2 


5541.4 


11 


45.829 


40.131 


42.535 


33 


7528.1 


6592.0 


6987.0 


12 


57.790 


50.604 


53.636 


34 


9495.6 


8314.9 


8813.1 


13 


72.949 


63.878 


67.706 


35 


11972 


10483 


11111 


14 


91.861 


80.439 


85.258 


36 


15094 


13217 


14009 


15 


115.86 


100.75 


107.53 


37 


19030 


16664 


17662 


16 


146.10 


127.94 


135.60 


38 


24003 


21018 


22278 


17 


184.26 


168.35 


171.02 


39 


30266 


26503 


28091 


18 


232.34 


203.45 


215.64 


40 


38176 


33342 


35432 



Note. — In this table the iron and copper employed are supposed to be 
nearly pure. The specific gravity or* the former was taken at 7.774 ; that of the 
latter, at 8.878. The specific gravity of the brass was taken at 8.376. 



WIRE AND WIRE GAUGES. 



47 



To find the number of feet in a pound of wire of any material not 
given in the table, of any size, American gauge, its specific 
gravity being known. 

Rule. — Multiply the number of feet in a pound of iron wire of 
the same size by 7.774, and divide the product by the specific grav- 
ity of the wire whose length is sought; or ordinarily, for steel wire, 
multiply the number of feet in a pound of iron wire of the same 
size by 0.991. 

To find the number of feet in a pound of wire of any given No., 
Birmingham gauge. 
Rule. — Multiply the number of feet in a pound of the same 
kind of wire, same No., American gauge, by the size, American 
gauge, and divide the product by the size, Birmingham gauge. 

Example. — In a pound of copper wire No. 16, American 
gauge, there are 127.94 feet : how many feet are there of the same 
kind of wire, same No., Birmingham gauge ? 

(127.94 X -05082) -^- .065 == 100.03. Am. 

To find the weight of any given length of wire of any given No. or 
size, American gauge, or the length in any given weight, by help 
of the foregoing table. 
Example. — Required the weight of 600 feet of No. 18 iron 

wire. 

600 -f- 232.34 = 2.5822 lbs. = 2 lbs. 9± oz., nearly. Am. 

Example. — Required the length in feet of 2± lbs. of No. 31 
brass wire. 

4394X2.5 = 10985. Am. 

Characteristics of Alloys of Copper and Zinc — Brass. 



Parts by Weight. 


Specific 
Gravity. 


Color. 


Denomination. 


Copper. 


Zinc. 


83 

80 

74£ 

66 

49^ 

33 


17 

20 

251. 

34 

501 

67 


8.415 
8.448 
8.397 
8.299 
8.230 
8.284 


Yellowish Red. 
it a 

Pale yellow. 
Full 

u tt 

Deep " 


Bath Metal. 
Dutch Brass. 
Rolled Sheet Brass. 
English Sheet Brass. 
German Sheet Brass. 
Watchmaker's Brass. 



Note. — To alloys of copper and zinc, generally, there is added a small 
quantity of lead, which renders them the better adapted for turning, planing, 
or tiling ; and, for the same reason, to alloys of copper and tin, there is usually 
added a small quantity of zinc (see Alloys and Compositions). 



TABLE 



Showing the Weight of One Square Foot of Rolled Metals, thickness 
corresponding to Nos., American Wire-gauge. 



Thickness. 


Iron. 


Steel. 


Copper. 


Brass. 


Lead. 


Zinc. 


JVo. 


Pounds. 


Pounds. 


Pounds. 


Pounds. 


Pounds. 


Pounds. 


1 


10.849* 


10.999 


13.109 


12.401 


17.102 


10.833 


2 


9.6611 


9.7953 


11.674 


11.043 


15.228 


9.6466 


3 


8.6032 


8.7227 


10.396 


9.8340 


13.562 


8.5903 


4 


7.6616 


7.7680 


9.2578 


8.7576 


12.078 


7.6501 


5 


6.8228 


6.9175 


8.2442 


7.7988 


10.755 


6.8126 


6 


6.0758 


6.1601 


7.3416 


6.9450 


9.5779 


6.0667 


7 


5.4105 


5.4856 


6.5377 


6.1845 


8.5291 


5.4024 


8 


4.8184 


4.8853 


5.8222 


5.5077 


7.5957 


4.8112 


9 


4.2911 


4.3507 


5.1851 


4.9050 


6.7645 


4.2847 


10 


3.8209 


3.8740 


4.6169 


4.3675 


6.0233 


3.8151 


11 


3.4028 


3.4501 


4.1117 


3.8896 


5.3642 


3.3977 


12 


3.0303 


3.0720 


3.6616 


3.4638 


4.7770 


3.0257 


13 


2.6985 


2.7360 


3.2607 


3.0845 


4.2539 


2.6934 


14 


2.4035 


2.4365 


2.9042 


2.7473 


3.7889 


2.3999 


15 


2.1401 


2.1698 


2.5829 


2.4463 


3.3737 


2.1369 


16 


1.9058 


1.9322 


2.3028 


2.1784 


3.0043 


1.9029 


17 


1.6971 


1.7207 


2.0506 


1.9399 


2.6753 


1.6945 


18 


1.5114 


1.5324 


1.8263 


1.7276 


2.3826 


1.5091 


19 


1.3459 


1.3646 


1.6263 


1.5384 


2.1217 


1.3439 


20 


1.1985 


1.2152 


1.4482 


1.3700 


1.8893 


1.1967 


21 


1.0673 


1.0821 


1.2897 


1.2300 


1.6768 


1.0657 


22 


.95051 


.96371 


1.1485 


1.0865 


1.4984 


.94908 


23 


.84641 


.85815 


1.0227 


.96749 


1.3343 


.84514 


24 


.75375 


.76422 


.91078 


.86158 


1.1882 


.75262 


25 


.67125 


.68057 


.81109 


.76728 


1.0582 


.67024 


26 


.59775 


.60605 


.72228 


.68326 


.94229 


.59685 


27 


.53231 


.53970 


.64345 


.60846 .83913 


.53151 


28 


.47404 


.48062 


. .57280 


.54185 


.74728 


.47333 


29. 


.42214 


.42800 


.51009 


.48242 


.66546 


.42151 


30 


.37594 


.38116 


.45426 


.42972 


.59263 


.37538 



Note. — In calculating the foregoing table, the specific gravities were 
taken as follows: viz., iron, 7.200; steel, 7.300; copper, 8.700; brass, 8.230; 
lead, 11.350; Zinc, 7.189. 



TIN PLATES. 



49 



TIN PLATES. 



Brand 

Marks. 



IC 

IC 

HC 

HX 

IX 

IXX 

IXXX 

IXXXX 

IX 

IXX 

DC 

DX 

DXX 

DXXX 

DXXXX 

SDC 

SDX 



Size of 


No. of 


Net 


Sheets in 


Sheets 


Weight 


Inches. 


in Box. 


in lbs. 


14 X 14 


200 


140 


14 X 10 


225 


112 


14 X 10 


225 


119 


14 X 10 


225 


147 


14 X 10 


225 


140 


14 X 10 


225 


161 


14 X 10 


225 


182 


14 X 10 


225 


203 


14 X 14 


200 


174 


14 X 14 


200 


200 


17X12^ 


100 


105 


17X12^ 


100 


126 


17X121 


100 


147 


17X12| 


100 


168 


17X12^ 


100 


189 


15XH 


200 


168 


15XH 


200 


189 



Brand Marks. 



Size of 
Sheets in 
Inches. 



15 X 
15 X 
15 X 
14 X 
12X 



SDXX 
SDXXX 
SDXXXX 
TT 

IC 

1X12 X 

IXXJ12X 

IXXX! 12 X 

IXXXXll2 X 

IC!-20 X 

IX!20 X 

IXX20 X 

IXXX20 X 

IXXXX 20 X 

Ternes ICl20 X 

IXJ20 X 



No. of 
1 Sheets 
in Box. 



200 
200 
200 
225 
225 
225 
225 



12 225 



14| 112 
14 

14 
14 
14 
14 
14 



112 
112 
112 
112 
112 
112 



Net 
Weight 
in lbs. 



210 
231 
252 
112 
119 
147 
168 
189 
210 
112 
140 
161 
182 
203 
112 
140 



Note. — The above table includes all the regular sizes and qualities of 
tin plates, except " wasters." Other sizes, such as 10 X 10, 11 X H» 13 X 13, 
&c, of the different brands, are often imported into the United States to 
order. 

Common English Sheet Iron, Nos. 10 to 28, Birmingham gauge, 
widths from 24 to 36 inches. 

R. G. Sheet Iron, Nos. 10 to 30, Birmingham gauge, widths from 
24 to 36 inches. 

American Puddled Sheet Iron, Nos. 22 to 28, Birmingham 
gauge, widths from 24 to 36 inches. 

Russia Sheet Iron, Nos. 16 to 8 inclusive, Russia gauge, sheets 
28 X 56 inches. 

Sheet Zinc, Nos. 16 to 8, Liege gauge, widths from 24 to 40 
inches ; length 84 inches. 

Copper Sheathing, 14 X 48 inches, 14 to 32 oz. (even numbers), 
per square foot. 

Yellow Metal, in sheets, 48 X 14 inches, 14 to 32 oz. (even num- 
bers), per square foot. 
5 



TABLE 

Showing the Capacity, in Wine Gallons, of Cylindrical Cans, of 
different diameters, at One Inch depth. Diameter in Inches. 



Diam'r. 


Gallons. 


Diam'r. 


Gallons. 


Diam'r. 


Gallons. 


Diam'r. 


Gallons. 


inches. 




inches. 




inches. 




inches. 




6 


.1224 


m 


.5102 


184 


1.164 


24| 


2.083 


64 


.1328 


m 


.5313 


18| 


1.195 


25 


2.125 


64 


.1437 


12| 


.5527 


19 


1.227 


254 


2.167 


6| 


.1549 


13 


.5746 


194 


1.260 


254 


2.211 


7 


.1666 


13| 


.5969 


1% 


1.293 


251 


2.254 


n 


.1787 


m 


.6197 


19| 


1.326 


26 


2.298 


74 


.1913 


13| 


.6428 


20 


1.360 


264 


2.343 


71 


.2042 


14 


.6664 


204 


1.394 


264 


2.388 


8 


.2176 


144 


.6904 


204 


1.429 


261 


2.433 


84 


.2314 


144 


.7149 


20| 


1.464 


27 


2.479 


84 


.2457 


141 


.7397 


21 


1.499 


274 


2.524 


81 


.2603 


15 


.7650 


214 


1.535 


274 


2.571 


9 


.2754 


154 


.7907 


214 


1.572 


271 


2.518 


94 


.2909 


15i 


.8169 


211 


1.608 


28 


2.666 


9h 


.3069 


15| 


.8434 


22 


1.646 


284 


2.713 


91 


.3233 


16 


.8704' 


224 


1.683 


284 


2.762 


10 


.3400 


164 


.8978 


224 


1.721 


281 


2.810 


104 


.3572 


164 


.9257 


221 


1.760 


29 


2.859 


104 


.3749 


161 


.9539 


23 


1.799 


294 


2.909 


101 


.3929 


17 


.9826 


234 


1.837 


291 


3.009 


11 


.4114 


174 


1.0120 


234 


1.877 


30 


3.060 


111 


.4303 


17i 


1.0410 


231 


1.918 


304 


3.163 


111 


.4497 


171 


1.0710 


24 


1.958 


31 


3.264 


111 


.4694 


18 


1.1020 


244 


1.999 


314 


3.374 


12 


.4896 


184 


1.1320 


244 


2.041 


32 


3.482 



Applications of the foregoing table. 

Example. — A cylindrical can is 114 inches in diameter, and its 
depth is 18 f inches ; required its capacity. 

.4303 X 18f = 8 gallons. Ans. 

Example. — The diameter of a can containing oil is 264 inches, and 
the oil is 144 inches in depth. How many gallons are there of the oil ? 

2.388 X 144 = 34.6 gallons. Ans. 

Example. — A can is to be constructed that will hold just 36 gal- 
lons, and its diameter is to be 18 inches ; what must be its depth? 

36-1-1.102 = 32| inches. Ans. 



CAPACITY OF CYLINDRICAL CANS. 51 

Example. — A cylindrical can is to be constructed that shall have 
a depth of 15 inches and a capacity of just 5 gallons ; what must be 
its diameter? 

5 -J- 15 = .3333 = capacity of can in gallons for each inch of depth ; 
and against .3333 gallon in the table, or the quantity in gallons 
nearest thereto, is 10 inches, the required, or nearest tabular diam- 
eter. Ans. 

Note. — The table is not intended to meet demands of the nature of the one contained in 
the last example, with accuracy, unless the fractional part of the diameter, if there be a 
fractional part, is £, £ or % inch. As, however, the diameter opposite the tabular gallon 
nearest the one sought, even at its greatest possible remove, can be but about £ inch from 
the diameter required, we can, by inspection, determine the diameter to be taken, or true 
answer to the inquiry, sufficiently near for practical purposes, be the fraction what it may. 
Or, to throw the demand into a mathematical formula : As the tabular gallon nearest the 
one sought is to the diameter opposite, so is the tabular gallon required to the required 
diameter, nearly. Thus, in answer to the last query, 

.3400 : 10 : : 3333 : 9.8 inches, the required or true diameter, nearly. 

For a mathematical formula strictly applicable to this question, see Gauging 

Or, for a formula more strictly geometrical, we have 

. Capacity X 231 

a/ -^ ~r,r, = diameter. 

^ Depth X -7854 

The true diameter, therefore, for the supposed can, is 

. 231X5 



15 X -7854 



= 9.9— inches. 



52 WEIGHT OF PIPES. 

WEIGHT OF PIPES. 

The weight of one foot in length of a pipe, of any diameter 
and thickness, may be ascertained by multiplying the square of its 
exterior diameter, in inches, by the weight of 12 cylindrical inches of 
the material of which the pipe is composed, and by multiplying the 
square of its interior diameter, in inches, by the same factor and sub- 
tracting the product of the latter from that of the former, — the 
remainder or difference will be the weight. This is evident from the 
fact that the process obtains the weight of two solid cylinders of equal 
length, (one foot,) the diameter of one being that of the pipe, and the 
other that of the vacancy, or bore. For very large pipes, the dimen- 
sions may be taken in feet, and the weight of a cylindrical foot of the 
material used as the factor, or multiplier, if desired. 

The weight of 12 cylindrical inches (length 1 foot, diameter 1 inch) 
of 

Malleable Iron = 2.6543 lbs. 

Cast Iron =2.4573 " 

Copper, wrought, = 3.0317 " 

Lead " =3.8697 " 

Cast Iron— 1 cyl. foot— = 353.86 " 

Therefore — Example. — Required the weight of a copper pipe 
whose length is 5 feet, exterior diameter 3^ inches, and interior 
diameter 3 inches. 

U = -^ X -T- = 10.5625 X 3.0317 = 32.022 -f- 
3 X 3 = 9 X 3.0317 = 27.285 -\- 

Ans. 4.737 X 5 = 23.685 lbs. 

Example. — Required the weight of a cast iron pipe, whose length 
is 10 feet, exterior diameter 38 inches, and interior diameter 3 feet. 
38 2 X 2.4573 — 36 2 X 2.4573 = 363.68 X 10 = 3636.8 lbs. Ans. 



Or, 38* — 36 2 = 148 X 2.4573 = 363.68 X 10 = 3636.8 lbs. Ans. 

Example. — Required the weight of a lead pipe, whose length is 
1200 feet, exterior diameter £ of an inch, and interior diameter T 9 B 
of an inch. 

I X I — 1| = .765625, and T \ X * = AV = -316406, and 
.765625 — .316406 = .449219 X 3.8697 X 1200 = 2086 lbs. Ans. 

Example. — The length of a cast-iron cylinder is 1 foot, its 
exterior diameter is 12 inches, and its interior diameter 10 inches ; 
required its weight. 

12 2 _ 10 2 -=■■ 44 X 2.4573 = 108. 12 lbs. Ans. 
Or, 144 : 353.86 : : 44 : 108.12 lbs. Ans. 



WEIGHT OF PIPES. 



53 



The following Table exhibits the coefficients of wdght, in pounds, oj 
one foot in length, of various thicknesses, of different kinds of pipe, oj 
any diameter whatever. 





Thickness 
in Inchos. 


Wrought 
Iron. 


Copper. 


Lead. 






■h 


.332 


.379 


.484 






1 
TF 


.664 


.758 


.9675 






A 


.995 


1.137 


1.451 






i 


1.327 


1.516 


1.935 






A 


1.658 


1.894 


2.417 






3 
TF 


1.99 


2.274 


2.901 






7 
"3~2" 


2.323 


2.653 


3.386 






1 
4~ 


2.654 


3.032 


3.87 






A 


3.318 


3.79 


4.837 






1 


3.981 

CAST 


4.548 

IRON. 


5.805 








Thickness. 


Factor. 


Thickness. 


Factor. 


Thickness. 


Factor. 


3 
TB" 


1.842 


6.143 


11 


12.287 


1 

4 


2.457 


4* 


7.372 


u 


14.744 


f 


3.686 


7 
8" 


S.f: 


1} 


17.201 


I 
2 


4.901 


1 


9.829 


2 


19.659 



To obtain the weight of pipes by means of the above Table — 
Rule. — Multiply the diameter of the pipe, taken from the interior 
surface of the metal on the one side to the exterior surface on the 
opposite, (interior diameter -f- thickness,) in inches, by the number 
in the table under the respective metal s name, and opposite the 
thickness corresponding to that of the pipe — the product will be the 
weight, in pounds, of one foot in length of the pipe, and that product 
multiplied by the length of the pipe, in feet, will give the weight for 
any length required. 

Example. — Required the weight of a copper pipe whose length is 
5 feet, interior diameter and thickness 3£ inches, and thickness £ of an 
inch. 

3 J = - 2 F 5 - = 3.125 X 1.516 X 5 = 23.687 lbs. Am. 

Example. — Required the weight of a cast iron pipe, 10 feet in 
length, whose interior diameter is 3 feet, and whose thickness is 1 inch. 
36 -f- 1 = 37 X 9.829 X 10 = 3636.73 lbs. Ans. 
5* 



54 WEIGHT OF BALLS AND SHELLS. 

WEIGHT OF CAST IRON AND LEAD BALLS. 

To find the weight of a sphere or globe of any material — 
Rule. — Multiply the cube of the diameter, in inches, or feet, by 
the weight of a spherical inch or foot of the material. 
The weight of a spherical inch of 

Cast Iron . = .1365 lbs. 
Lead . . = .215 " 
Therefore — Example. — Required the weight of a leaden ball 
whose diameter is ^ of an inch. 

|XiX|=6l= .015625 X 215 = .00336 lb. Ans. 
Example. — Required the weight of a cast iron ball whose diameter 
•is 8 inches. 

8 3 X -1365 = 69.888 lbs. Ans. 

Example. — How many leaden balls, having a diameter I of an inch 
each, are there in a pound ? 

1 -7- .00336 = - 1 -°i 3°F -- = 298 - Ans - 
Example. — What must be the diameter of a cast iron ball, to 
weigh 69.888 lbs? 

69.888 -J- .1365 = ^/512 = 8 inches. Ans. 

Example. — What must be the diameter of a leaden ball to equal 
in weight that of a cast iron ball, whose diameter is 8 inches? 
[Lead is to cast iron as .215 to .1365, as 1.575 to 1.] 
8 3 = 512 -r- 1.575 = ^/325 = 6.875 inches. Ans. 



WEIGHT OF HOLLOW BALLS OR SHELLS. 

The weight of a hollow ball is the weight of a solid ball of the 
same diameter, less the weight of a solid ball whose diameter is that 
of the interior diameter of the shell. 

Example. — Required the weight of a cast iron shell whose ex- 
terior diameter is 6^ inches, and interior diameter 4| inches. 

6£ =^ 5 - X -\ 5 - X 2 -i- = 244.14 X -1365 = 33.33 

4| =4.25 3 X -1365 = 10.48 

22.85 lbs. Ans. 

Or, If we multiply the difference of the cubes, in inches, of the two 
diameters — the exterior and interior — by the weight of a spherical 
inch, we shall obtain the same result. 

Example. — Required the weight of a cast iron shell whose ex- 
terior diameter is 10 inches and interior diameter 8 inches. 
10 3 — 8 3 X .1365 = 68.612 lbs. Ans. 



ANALYSIS OF COALS. 



55 



ANALYSIS OF COALS. 



Description. 

Breckinridge, Ky., 
"Albert," N. B., 
Chippenville, Pa., 
Kanawha, " 
Pittsburg, " 
Cannel, sp. gr. 1.4 
Newcastle, 
Cumberland, 
Anthracite, a'v'g., 



Volatile Matter. 


Carbon. 


62.25 


29.10 


61.74 


32.14 


49.80 




41.85 




32.95 




35.28 


64.72 


24.72 


75.28 


18.40 


80. 


3.43 


89.46 



Ash. 

8.65 
6.12 



1.60 
7.11 



Woods of most descriptions vary little from 80 per cent, volatile 
matter, and 20 per cent, charcoal. 

Table — Exhibiting the Weights, Evaporative Powers, fyc, of Fuels, 
from Report of Professor Walter R. Johnson. 







Weight 


Lbs. of Water 
at 212 degrees 


Lbs. of Water 


Weight of 


Designation of Fuel. 


Specific 
Grav- 


per 
Cubic 


converted into 
Sieam by I 


at 212 degrees 
converted into 


Clinkers 
from 100 lbs. 




ity. 


Foot. 


Cubic Foot of 
Fuel. 


Steam by 1 lb. 
of Fuel. 


of Coal. 


Anthracite Coals. 










Beaver Meadow, No. 3 


1.610 


54.93 


526.5 


9.21 


1.01 


Beaver Meadow, No. 5 


1.554 


56.19 


572.9 


9.88 


.60 


Forest Improvement 


1.477 


53.66 


577.3 


10.06 


.81 


Lackawanna 


1.421 


48.89 


493.0 


9.79 


1.24 


Lehigh 


1.590 


55.32 


515.4 


8.93 


1.08 


Peach Mountain 


1.464 


53.79 


581.3 


10.11 


3.03 


Bituminous Coals. 












Blossburgh 


1.324 


53.05 


522.6 


9.72 


3.40 


Cannelton, la. 


1.273 


47.65 


360.0 


7.34 


1.64 


Clover Hill 


1.285 


45.49 


359.3 


7.67 . 


3.86 


Cumberland, average, 


1.325 


53.60 


552.8 


10.07 


3.33 


Liverpool 


1.262 


47.88 


411.2 


7.84 


1.86 


Midlothian 


1.294 


54.04 


461.6 


8.29 


8.82 


Newcastle 


1.257 


50.82 


453.9 


8.66 


3.14 


Pictou 


1.318 


49.25 


478.7 


8.41 


6.13 


Pittsburgh 


1.252 


46.81 


384.1 


8.20 


.94 


Scotch 


1.519 


51.09 


369.1 


6.95 


5.63 


Sydney 


1.338 


47.44 


386.1 


7.99 


2.25 


Coke. 












Cumberland 




31.57 


284.0 


8.99 


3.55 


Midlothian 




32.70 


282.5 


8.63 


10.51 


Natural Virginia 


1.323 


46.64 


407.9 


8.47 


5.31 


Wood. 












Dry Pine Wood 




21.01 


98.6 


4.69 





56 MENSURATION OF LUMBER. 

MENSURATION OF LUMBER. 

To find the contents of a board. 

Rule. — Multiply the length in feet by the width in inches, and 
divide the product by 12 ; the quotient will be the contents in square 
feet. 

Example. — A board is 16 feet long and 10 inches wide ; how 
many square feet does it contain ? 

16 X 10 = 160 -=-12 = 13 T 4 2 . Ans. 

To find the contents of a plank, joist, or stick of square timber. 

Rule. — Multiply the product of the depth and width in inches by 
the length in feet, and divide the last product by 12 ; the quotient is 
the contents in feet, board, measure. 

Example. — A joist is 16 feet long, 5 inches wide, and 2£ inches 
thick ; how many feet does it contain, board measure 1 

5 X 2.5 X 16 -J- 12 = 16 T V Ans. 

To find the solidity of a plank, joist, or stick of square timber. 

Rule. — Multiply the product of the depth and width in inches 
by the length in feet, and divide the last product by 144 ; the quo- 
tient will be the contents in cubic feet. 

Example. — A stick of timber is 10 by 6 inches, and 14 feet in 
length ; what is its solidity 1 

10 X 6 = 60 X 14 = 840 -$- 144 = 5f feet. Ans. 

Note. — If aboard, plank, or joist is narrower at one end than the other, 
add the two ends together and divide the sum by 2; the quotient will be the 
mean width. And if a stick of squared timber, who se solidity is required, is 
narrower at one end than the other (A + a + V AaJ -f- 3 = mean area. A and 
a being the areas of the ends. 

To measure round timber. 

Rule (in general practice.) — Multiply the length, in feet, by 
<rhe square of £ the girt, in inches, taken about £ the distance from 
the larger end, and divide the product by 144 ; the quotient is con- 
sidered the contents in cubic feet. For a strictly correct rule for 
measuring round timber, see Mensuration or Solids — Frustum of a 
Cone. 

Example. — A stick of round timber is 40 feet in length, and 
girts 88 inches ; what is its solidity ? 

88 -5- 4 = 22 X 22 = 484 X 40 = 19360 + 144 =» 134.44 cub. ft. Ans. 



MENSURATION OF LUMBER. 



57 



TJie following TABLE is intended to facilitate the measuring of Round 
Timber , and is predicated upon the foregoing Rule. 



i Girt in 


Area in 


i Girt in 


Area in 


i Girt in 


Area in 


i Girt in 


Area in 


Inches. 


Feet. 


Inches. 


Feet. 


Inches. 


Feet. 


Inches. 


Feet. 


6 


.25 


12 


1. 


18 


2.25 


24* 


4. 


H 


.272 


12| 


1.042 


I84 


2.313 


244 


4.084 


6-i 


.294 


m 


1.085 


m 


2.376 


24£ 


4.168 


61 


.317 


121 


1.129 


181 


2.442 


241 


4.254 


7 


.34 


13 


1.174 


19 


2.506 


25 


4.34 


n 


.364 


134 


1.219 


194 


2.574 


254 


4.428 


n 


.39 


m 


1.265 


m 


2.64 


25£ 


4.516 


71 


.417 


13,1 


1.313 


191 


2.709 


251 


4.605 


8 


.444 


14 


1.361 


20 


2.777 


26 


4.694 


H 


.472 


144 


1.41 


204 


2.898 


264 


4.785 


8£ 


.501 


14£ 


1.46 


20£ 


2.917 


26£ 


4.876 


81 


.531 


141 


1.511 


201 


2.99 


261 


4.969 


9 


.562 


15 


1.562 


21 


3.062 


27 


5.062 


H 


.594 


15_i 


1.615 


214 


3.136 


274 


5.158 


H 


.626 


15£ 


1.668 


21£ 


3.209 


27h 


5.252 


91 


659 


151 


1.722 


211 


3.285 


271 


5.348 


10 


.694 


16 


1.777 


22 


3.362 


28 


5.444 


104 


.73 


164 


1.833 


224 


3.438 


284 


5.542 


10£ 


.766 


16£ 


1.89 


22£ 


3.516 


28£ 


5.64 


101 


.803 


161 


1.948 


221 


3.598 


281 


5.74 


n 


.84 


17 


2.006 


23 


3.673 


29 


5.84 


HI 


.878 


174 


2.066 


234 


3.754 


294 


5.941 


m 


.918 


m 


2.126 


23£ 


3.835 


29£ 


6.044 


Hi 


.959 


171 


2.187 


231 


3.917 


30 


6.25 



To find the solidity of a log by help of the preceding table. 

Rule. — Multiply the tabular area opposite the corresponding 
4 girt, by the length of the log in feet, and the product will be the 
solidity in feet. 

Example. — The 4 girt of a log is 22 inches, and the length of the 
log is 40 feet ; required the solidity of the log. 

3.362 X 40 = 134.48 cubic feet. Ans. 

Note. — Though custom has established, in a very general way, the preceding method 
as that whereby to measure round timber, and holds, in most instances, the solidity to be 
that which the method will give, there seems, if the object sought be the real solidity 
of the stick, neither accuracy, justice, nor certainty, in the practice. 

Thus, in the preceding example, the stick was supposed to be 40 feet in length, and 88 
inches in circumference at £ the distance from the larger end, and was found, by the 
method, to contain 134.44 cubic feet : now 88 -r 3.1416 = 28 inches, = the diameter at i 
the distance from the greater base, and retaining this diameter and the length, we may 



58 MENSURATION OP LUMBER. 

suppose, with sufficient liberality, and without being far from the general run of such 
sticks, the diameter at the greater base to be 30 inches, and that of the less to be 21 
inches, and — 
By a correct rule the stick contains — 

30 X 24 = 720 -j- 12 = 732 X -7854x40=22996-^144 = 159.7 cubic feet, or 19 per 
cent, more than given by the method under consideration ; and we need hardly add that 
the nearer the stick approaches to the figure of a cylinder, the wider will be the difference 
between the truth »nd the result obtained by the method referred to. Thus, suppose the 
stick a cylinder, 28 inches in diameter, and 40 feet in length ; and we have, by the falla- 
cious rule, as above, 134.44 cubic feet ; and — 

By a correct method, we have — 

28 2 X .7854X40 = 24630-^-144 =171 cubic feet, or over 27 per cent, more than fur- 
nished by the erroneous mode of practice. 

Again : suppose the stick in the form of a cone, 30 inches at the base, and tapering to a 
point at 150 feet in length ; and we have, by a correct rule — 

30 2 -^- 3 = 300 X -7854 X 150 = 35343 -r 144 = 245.44 cubic feet; and by the ordinary 
method of gauging, or the aforementioned practice, we have — 

20 X 3.1416 = 62.832 -r 4 = 15.70$ 2 X 150 = 37011.19 ^- 144= 257 cubic feet, or nearly 
i§ per cent, more than the stick actually contains. 

In short, without taking into account anything for the thickness of the bark, that may 
be supposed to be on the stick, the method is correct only when the stick tapers at the 
rate of 5 j- inches diameter per each 10 feet in length, or over £ inch diameter to each foot 
in length of the stick. 

If, however, we suppose the stick as before, (30 inches at the greater base, 24 inches at 
the smaller, and 40 feet in length,) and suppose the bark upon it to be 1- inch thick, we 
shall have, by the usual method, 134.44 cubic feet, as before. And, exclusive of the bark, 
b y a corr ect metho d, we shall have. 

30 — 2 X 24 — 2 = 616 -4- 12 = 628 X -7854 X 40 = 19729 -f- 144 = 137 cubic feet, or 
only about 2 per cent, more than that furnished us by the usual practice. 

The following simple rule for measuring round timber is suffi- 
ciently correct for most practical purposes : — 

Rule. — Multiply the square of one-fifth of the mean girt, (exclu- 
sive of bark,) in inches, by twice the length of the stick in feet, and 
divide the product by 144 ; the quotient will be the solidity in feet. 

To find the solidity of the greatest rectangular stick that may he cut 
from a given log, or from a stick of round timber of given dimen- 
sions. 

Rule. — Multiply the square of the mean diameter of the log, in 
inches, by half the length of the log, in feet, and divide the product 
by 144. 

Example. — The diameter (exclusive of bark) of the greater base 
of a stick of round timber is 30 inches, and that of the less base is 
24 inches, and the stick is 40 feet in length ; required the solidity 
of the greatest rectangular stick that may be cut from it. 

30 X 24 + 1 (30 — 24) 2 = 732 = square of mean diameter,* and 

732 X 20 = 14640 -j- 144 = 101| cubic feet. Ans. 

* Except in the case of a cylinder, there is a difference betwixt the mean diameter of a 
solid having circular bases, and the middle diameter of that solid. The mean diameter 
reduces the solid to a cylinder ; the middle diameter is the diameter midway between the 
two bases. 



MENSURATION OF LUMBER. 59 

Note. — The foregoing stick will make — 

14640 -^ 16 = 915 feet of square-edged boards 1 inch thick ; 
Or, 101S X 9=915. 

To find the solidity of the greatest square stick that may be cut from a 
given log, or from a stick of round timber of given dimensions. 

Rule. — Multiply the square of the diameter of the less end of the 
log, in inches, by half the length of the log, in feet, and divide 
the product by 144. 

Example. — The preceding supposed log will make a square stick 
containing — 

242 x 4^o_ = 1152 _i. 144 _ 80 cubic feet. 

Diameter multiplied by .7071 = side of inscribed square. 

To find the contents, in Board Measure, of a log, no allowance being 
made for wane or saw-chip. 

Rule. — Multiply the square of the mean diameter, in inches, by 
the length in feet, and divide the product by 15.28. 

Or, Multiply the square of the mean diameter in inches, by the 
length in feet, and that product by .7854, and divide the last prod- 
uct by 12. 

The cubic contents of a log multiplied by 12, equal the contents 
of the log, board measure. 

The convex surface of a Frustum of a Cone = (C -j- c) X h slant 
length ; C being the circumference of the greater base, and c the 
circumference of the less. 



60 GAUGING. 



GAUGING. 



Rules for finding the capacity in gallons or bushels of different 
shaped Cisterns, Bins, Casks, <Sfc, and also, by way of examples, for 
constructing them to given capacities. 

Rule — 1. When the vessel is rectangular. Multiply the interior 
length, breadth, and depth, in feet together, and the product by the 
capacity of a cubic foot, in gallons or bushels, as desired for its 
capacity. 

Rule — 2. When the vessel is cylindrical. Multiply the square of 
its interior diameter in feet, by its interior depth in feet, and the prod- 
uct by the capacity of a cylindrical foot in gallons or bushels, as 
desired for its capacity. 

Rule — 3. When the vessel is a rhombus or rhomboid. Multiply 
its interior length, in feet, its right-angular breath in feet, and its 
depth in feet together, and the product by the capacity of a cubic foot 
in the special measure desired for its capacity. 

Rule — 4. When the vessel is a frustum of a cone — a round vessel 
larger at one end than the other, whose bases are planes. Multiply 
the interior diameter of the two ends together, in feet, add £ the 
square of their difference in feet to the product, multiply the sum by 
the perpendicular depth of the vessel in feet, and that product by the 
capacity of a cylindrical foot in the unit of measure desired for its 
capacity. 

Rule — 5. When the vessel is a prismoid or the frustum of any 
regular pyramid. To the square root of the product of the areas of its 
ends in feet, add the areas of its ends in feet, multiply the sum by 
£ its perpendicular depth in feet, and that product by the capacity of 
a cubic foot in gallons or bushels, as desired for its capacity. 

If it is found more convenient to take the dimensions in inches, do 
so ; proceed as directed for feet, divide the product by 1728, and mul- 
tiply the quotient by the capacity of the respective foot as directed. 
Or, multiply the capacity in inches by the capacity of the respective 
inch in gallons or bushels ; — by the quotient obtained by dividing the 
capacity of the respective foot in gallons or bushels by 1728 — for 
the contents. 

Rule — 6. When the vessel is a barrel, hogshead, pipe, dfc. Mul- 
tiply the difference in inches between the bung diameter and head 
diameter, (interior,) if the staves be 

much curved, . . by .7 "1 

medium curved, ' . by .65 I See 6g . 

straighter than medium, by .6 ( ^ ° 
nearly straight, . by .55 J 

and add the product to the head diameter, taken in inches ; then mul- 
tiply the square of the sum by the length of the cask in inches, and 
divide the product by the capacity in cylindrical inches of a gallon or 



GAUGING. 61 

bushel as desired for the contents. Or, divide the contents in cylin- 
drical inches, as above found, by 1728, and multiply the quotient by 
the capacity of a cylindrical foot in gallons or bushels as desired for 
its contents. Or, multiply the capacity in cylindrical inches by the 
capacity of a cylindrical inch, in gallons or bushels, as desired, — 
that is, by the quotient obtained by dividing the capacity of a cylin- 
drical foot in gallons or bushels, by 1728, for the contents. 
The capacity of a 



cubic foot = 

7.4805 Winchester wine gallons. 

6.1276 Ale 

6.2321 Imperial " 

.80356 Winchester bushel. 

.62888 » heaped " 

.64285 " l£ even " 

.779 Imperial " " 



CYLINDRICAL FOOT = 

5.8751 • Winchester wine gallons. 

4.8126 Ale 

4.8947 Imperial " 

.63111 Winchester bushel. 

.49391 " heaped " 

.50489 " li even «• 

.61183 Imperial 



Example. — Required the capacity in Winchester bushels of a 
rectangular bin, whose interior length is 12 feet, breadth 6 feet, and 
depth 5 feet. 

12 X 6 X 5 X -8035 = 289.26 bushels. Ans. 
Example. — Required the capacity in Winchester wine gallons of 
a cylindrical can, whose interior diameter is 18 inches, and depth 3 
feet. 

18 X 18 X 36X 5.875 -r- 1728 = 39.66 gallons. Ans. 
Or, 1.5 X 1-5 X 3 X 5.875 = 39.66 gallons. Ans. 

Or, 18 X 18 X 36 X -0034 = 39.66 gallons. Ans. 

Example. — How many Winchester bushels in 39.66 wine gal- 
lons? 

39.66 X .10742 = 4.26 bushels. Ans. 

Example. — How many wine gallons in 4.26 Winchester bushels ? 
4.26 X 9.3092 = 39.66 gallons. Ans. 

Example. — How many wine gallons will a cistern in the form of 
a frustum of a cone hold, having the interior diameter of one of its 
ends 6 feet, and that at the other 8 feet, and its perpendicular depth 
9 feet? 

8 — 6 = 2, and 2 2 -f- 3 = 1.333 = \ square of dif. of diameters, and 
6 X 8 + 1.333 = 49.333 X 9 X 5.8751 = 2608.55 gals. Ans. 
Or, 6 X 8 4-8 2 -[- 6^ = 148 Xf X 5.8751 = 2608.55 gals. Ans. 
Or, (83 _ 6* ) -i- (8 — 6) = 148 X f X 5.8751 = 2608.55 gals. Ans. 
Or, 96 — 72 = 24 and (24 2 -r- 3) = 192, and 

96 X 72 + 192 = 7104 X108 X .0034 = 2008.55 gals. Ans. 
6 



62 GAUGING. 

Example. — What is the capacity in Winchester bushels of a cis- 
tern whose form is prismoid, the dimensions (interior) of one end 
being 8 by 6 feet, of the other 4 by 3 feet, and its perpendicular 
depth 12 feet 1 

8 X 6 = 48 = area of one end, and 4 X 3 = 12 = area of the other 
end ; then — 

48 X 12 = V 576 = 24 + 48 + 12 = 84 X"V 2 " X .80356 = 270 bush- 
els. Ans. 
Or, (8-J-4) -r-2 = 6,andX6-f-3) _j_2 = 4.5 = mean sectional areas 
of ends, and 

6X4. 5X4 = 4 area of mean perimeter, then 
8 X 6 + 4 X 3 + 6 X 4.5 X 4 = 168 X -V 2 - X -80356 = 270 bus. Ans. 

Example. — What must be the depth of a rectangular bin whose 
length is 12 feet, and breadth 6 feet, to hold 289.26 bushels ? 
289.26 -H (12 X 6 X .80356) = 5 feet. Ans. 

Example. — A cylindrical can, whose depth is to be 36 inches, is 
required to be made that will hold 40 gallons ; what must be the 
diameter of the can ? 

40 -J- (3 X 5.8751) = V2.27 = 1.506 feet. Ans. 
Or, 40 -J- (36 X .0034) = >/326.8 = 18.07 inches. Ans. 

Example. — A cylindrical can, whose interior diameter is to be 18 
inches, is required that will hold 40 gallons; what must be the 
interior depth of the can 1 

40 -h (18 2 X .0034) = 36.31 inches. Ans. 
Or, 40 -f- (1.5 s X 5.8751) = 3.026 feet. Ans. 

Example. — A cistern is to be built in the form of a frustum of a 
cone, that will hold 1800 gallons, and the diameter of one of its 
ends is to be 5 feet, and that of the other 7£ feet ; what must be the 
depth 1 

7.5 — 5 = 2.5, and 2.5 2 -f- 3 = 2.0833 = } square of difference of 
diameter, and 

1800 -f- (7.5 X 5 -f 2.0833) X 5.8751 = 7.74 feet. Ans. 
/7.5 X 5 + 7.52 + 52 \ 

Or, 1800 +{ -3 X 5.8751 )= 7.74 feet. Ans. 

Example. — The form, capacity, depth, and diameter of one end 
being determined on, and being as above, what must be the diameter 
of the other end ? 

c 
yr — %d 2 = y, c being the solidity in cylindrical measurement, A 



GAUGING. 63 

the depth, d the diameter of the given end or base, and y a quantity 
the square root of which is the sum of the required base and half the 
given base ; then 

1800 H- 5.8751 = 306.378 = solidity in cylindrical feet, and 

306.378 -H 7 -i A = 118.75 — (5 2 + # ) = V100 = 10 — f = 7.5 
feet. Ans. 

Example. — A measure is to be built in the form of a frustum of a 
cone, that will hold exactly 1 wine gallon, and the diameter of 
one of its ends is to be 4 inches, and that of the other 6 inches ; 
what must be its depth ? 

l4-(6X4 + H)X .0034 = 11.61 inches. Ans. 

231 6 X 4 -f 6 2 + 4 2 
Or, nozA -f- o = 11.61 inches. Ans. 

Example. — A measure in the form of a frustum of a cone holds 1 
wine gallon ; the diameter of one of its ends is 6 inches, and its 
depth is 11.61 inches ; what is the diameter of the other end? 

Jfic = 294.1176 -j- »- l iP- = 76 - (6 2 H- |) = V49 = 7 - f 

= 4 inches. Ans. 

CASK GAUGING. 

Cask-gauging, in a general sense, is a practical art, rather 
than a scientific achievement or problem, and makes no pretensions 
to strict accuracy with regard to the conclusions arrived at. The 
aim is, by means of a few satisfactory measurements taken of the 
outside, and an estimate of the probable mean thickness of the ma- 
terial of which the cask is composed (of which there must always 
remain some doubt), or by means of a few measurements taken of 
the inside, to determine, 1st, the capacity of the cask, and, 2d, the 
ullage, or capacity of the occupied or unoccupied space in a cask 
but partly full. And the Rule (Rule 6, page 60), which re- 
duces the supposed cask, or cask of supposed curvature, to a cylin- 
der, is as practically correct for the capacity of ordinary casks, as 
any rule, or set of rules, that can be offered for general purposes. 

Casks have no fixed form of their own, to which they severally 
and collectively correspond, nor are they in any considerable degree 
in conformity with any regular geometrical figure. 

Some casks — a few — those having their staves much curved 
throughout their entire length, are nearest in keeping with the 
middle frustum of a spheroid; others, slightly less curved than 
the preceding, correspond in a considerable degree to the middle 



64 GAUGING. 

frustum of a parabolic spindle; others, again — those having very 
little longitudinal curvature of stave to their semi-lengths — are 
nearly in keeping with the equal frustums of a paraboloid ; and others 
— a very few — those whose staves are straight from the bung diam- 
eter to the heads, or equal to that form, are in accordance with the 
equal frustums of a cone. 

The gauging rod, which is intended to be correct for casks of the 
most common form, gives for all casks, as may be seen in one of the 
following Examples, a solidity slightly greater (about 2£ per cent.) 
than would be obtained by supposing the cask in conformity with 
the third figure above alluded to. 

The Rule for finding the contents of a cask, by four dimensions, 
hereafter to be given, is intended as a general Rule for all casks, 
and, when the diameter midway between the bung and head can be 
accurately ascertained, will lead to a very close approach to the 
truth. 

From the length of a cask, taken from outside to outside of the 
heads, with callipers, it is usual to deduct from l'to 2 inches, to cor- 
respond with the thickness of the heads, according to the size of the 
cask, and the remainder is taken as the length of the interior. 

To the diameter of each head, taken externally, from i inch to 
Y^ inch should be added for common-sized barrels, -j^- inch for 40 gal- 
lon casks, and from £ inch to j^ inch for larger casks, to correspond 
with the interior diameters of the heads. 

If the staves are of uniform thickness, any sectional diameter of a 
cask may be nearly or quite ascertained, by dividing the circumfer- 
ence at that place by 3.1416, and subtracting twice the thickness of 
the stave from the quotient. «. 

For obtaining the diagonal of a cask by mathematical process, — 
the interior length, &c. &c. — see Rules, below. 

In the following formulas D denotes the bung diameter, d the 
head diameter, and I the length of the cask. 

The solidity of any cask is equal to its length multiplied by the 
square of its mean diameter multiplied by .7854. 

To calculate the contents of a cask from four dimensions. 

Rule. — To the square of the bung diameter add the square of 
the head diameter, and the square of double the diameter midway 
between the bung and head, and multiply the sum by ^ the length 
of the cask, for its cylindrical contents ; the product multiplied by 
.0034 expresses the contents in wine gallons. 

Example. — The length of the cask is 40 inches, its bung diameter 
28 inches, head diameter 20 inches, and the diameter midway be 



GAUGING. 65 

tween the bung and head is 25.6 inches ; how many gallons' capacity 
has the cask 1 

20 2 -j- 28 2 -f- 25.6 X 2* = 3805.44 X - 4 f°- X .0034 = 86.26 gals. Ans. 

(D^ -j- d 2 -f 2m 2 ) X & I X -7854 = cubic contents. 

J)2i^t2nf 

= square of mean diameter. 

By Rule 6, p. 68, this cask will hold — 

28 — 20 = 8 X .65 = 5.2 + 20 = 25.2 X 25.2 X 40 X -0034 = 86.36 
gallons. 

When the cask is in the form of the middle frustum of a spheroid. 
| D 2 -f- \d 2 = square of mean diameter. 

And a cask of this form, having the same head diameter, bung 
diameter, and length as the preceding, will hold — 

2X28?-{-20 2 x 40 x 0034 = 89 216 gallons, 
o 

When the cask is in the form of the middle frustum of a parabolic 
spindle. 

| D 2 -{- \ d 2 — T3- (D^ d) 2 = square of mean diameter. 

And a cask of this form, having the same head diameter, bung 
diameter, and length as the preceding, will hold — 

522| + 133|= 656 — 8.533 = 647.467 X 40 X .0034 == 88.055 gals. 

When the cask is in the form of two equal frustums of a paraboloid. 
£ D 2 -j-i d 2 = square of mean diameter. 

And a cask of this form, having the same head diameter, bung 
diameter, and length as the preceding, will hold — 

Y X 40 X -0034 = 80.51 gallons. 

When the cask is in the form of the equal frustums of a cone. 

£ D 2 -|- £ d 2 — -$ (D ^r d) 2 = square of mean diameter. 

Or,hD 2 + ld 2 +}T>d= " " " 

Or, DX^+MI>^) 2 = " " " 

And a cask of this form, having the same head diameter, bung 
diameter, and length as the preceding, will hold — 

28 X 20 + 21^ X40X-0034 == 79.06 gals. 
6 * 



66 GAUGING. 

To find the contents of a cask the same as would be given by the 
gauging rod. 

The gauging rod is constructed upon the principle that the cube 
of the diagonal of a cask, in inches, multiplied by biNt^ttj equals the 
contents of the cask, in Imperial gallons. 

The contents in wine gallons of either of the aforementioned 
casks, therefore, by the gauging rod, would be — 

31.241 3 X .0027 = 82£ gals. 

The decimal coefficient to take the place of .0027, for finding the 
contents of a cask in the form of the middle frustum of a spheroid 
= .002926 ; and for finding the contents of a cask in the form of the 
equal frustums of a cone = .002593. And between these extremes 
lies the decimal for other casks, or casks of intervening figures. 

To find the diagonal of a cask, when the interior is inaccessible. 

Rule. — From the bung diameter subtract half the difference of 
the bung and head diameters, and to the square of the remainder 
add the square of half the length of the cask, and the square root 
of the sura will be the diagonal. 

Example. — What is the diagonal of a cask whose bung diameter 
is 28 inches, head diameter 20 inches, and length 40 inches ? 

28- 20 = 8-^-2 = 4, and 28 — 4 = 24, then 

V (24 2 -f- 20 3 ) = 31.241 inches. Ans. 

To find the length of a cask, the head diameter, bung diameter and 
diagonal being given. 

( TTFd 2 \ 

V I diagonal 3 — D — — ^ — J —h 1 - 

And the interior length of a cask, whose interior head diameter, 
bung diameter and diagonal, are as the preceding, will be 

V (31.2412 — 242) = 20 X 2 = 40 inches. 

To find the solidity of a sphere. 
D 2 X I D X .7854 = cubic contents, D being the diameter. 

To find the solidity of a spherical frustum. 

( b 2 +d?\ 

I I h 2 ~f~ q J X ^ X .7854 = cubic contents, b and d being the 

bases, and h the height. 

Note. — For Rules in detail pertaining to the foregoing figures, and for ether figures, 
see Mensup.atio' of Solids. 



ULLAGE. 67 



ULLAGE. 



The ullage or wantage of a cask is the quantity the cask lacks of 
being full. 

To find the ullage of a standing cask, when the cask is half full or more. 

Rule. — To the square of the head diameter, add the square of 
the diameter at the surface of the liquor, and the square of twice 
the diameter midway between the surface of the liquor and the upper 
head, and divide the sum by 6 ; the quotient, multiplied by the 
distance from the surface of the liquor to the upper head, multiplied 
by .0034, will give the ullage in wine gallons. 

Example. — The diameters are as follows — at the upper head, 20 
inches ; at the surface of the liquor, 22 inches ; and at a point midway 
between these, 21± inches ; and the distance from the upper head 
to the surface of the liquor is 5 inches ; required the ullage. 



(20 2 -f- 22 2 -f 21.25 X2")-r6 = 448.37 X 5 X .0034 = 7.62 gal- 
lons. Ans. 

When the cask is standing, and less than half full, to find the ullage. 

Rule. — Make use of the bung diameter in place of the head 
diameter, and proceed in all respects as directed in the last Rule, 
and add the quantity found to half the capacity of the cask ; the 
sum will be the ullage. 

Example. — The bung diameter is 28 inches ; the diameter at the 
surface of the liquor, below the bung, is 26 inches ; the diameter 
midway between the bung and the surface of the liquor is 27.3 
inches ; and the distance from the surface of the liquor to the bung 
diameter is 5 inches ; required the quantity the cask lacks of being 
half full ; and also the ullage of the cask, its capacity being 86.26 
gallons. 



(282 -j-262 _|_ 27.3 X2)-r6 = 740.2 X 5 X -0034 = 12.58 gal- 
lons less than \ full. Ans. 
And, 86.26 -H 2 = 43.13 + 12.58 = 55.73 gallons ullage. Ans. 

When the cask is upon its bilge, and half full or more, to find the ullage. 

Rule. — Divide the distance from the bung to the surface of the 
liquor — (the height of the empty segment) — by the "whole bung 
diameter, and take the quotient as the height of the segment of a 
circle whose diameter is 1, and find the area of the segment; mul- 
tiply the area by the capacity of the cask, in gallons, and that 
product by 1.25 ; the last product will be the ullage, in gallons, as 



68 ULLAGE. 

found by the aid of the wantage-rod; and will be correct for casks 
of the most common form. 

Note. — The area of the segment of a circle = 

(ch'd k arc -f h ch'd £ arc + ch'd seg.) X height seg. X tV*> yer V nearl y- 
And, having the diameter of the circle and the height of the segment given, the chord 
of half the arc, and the chord of the segment may be found, thus — 

radius — height = cosine '; radius — cosine = sine 2 ; *>/ (sine 2 ) X 2 = ch?d of seg. 
sine 2 + height seg. 2 = ch'd h arc 2 , and \f (ch'd k arc2) = ch'd h arc. 

Example. — The bung diameter is 28 inches, the height of the 
empty segment 5.6 inches, and the capacity of the cask 86.26 gal- 
lons ; required the ullage of the cask, in gallons. 

5.6 — 28 = .2 = height of seg., diameter as 1. 

1 -j- 2 = .5 = radius. 

.5 — .2 = .3 = cosine. 

.5 2 — .3 2 = .16 = sine 2 , or square of half the base of the segment, 

/y/.16 = .4 X 2 = .8 = chord of segment, or base of segment. 

A 2 -f - -2 2 = .2 = square of chord of half the arc. 

V-2 = .4472 = chord of half the arc, then — 

.4472 -r- 3 = .1491, and .1491 -f-~4472 -f- J X 2 X T<r = -1H7, 
area of segment, and 

.1117 X 86.26 X 1-25 = 12 gallons. Ans. 

When the cash is upon its bilge, and less than half full, to find the 

ullage. 
Rule. — Divide the depth of the liquor by the bung diameter, and 
proceed in all respects as directed in the last Rule ; then subtract 
the quantity found from the capacity of the cask, and the difference 
will be the ullage of the cask. 

To find the quantity of liquor in a cask by its weight. 

Example. — The weight of a cask of proof spirits is 300 lbs., and 
the weight of the empty cask {tare) is 32 lbs. How many gallons 
are there of the liquor ? 

300 — 32 = 268 -j- 7.732 = 34| gallons. Ans. 

Customary Rule by Freighting Merchants, for finding the cubic meas- 
urement of casks. 

Bung diameter 2 X t length of cask = cubic measurement. 

Note. — One cubic foot contains 7.4805 wine gallons. 

* For several Rules in detail, for finding the area of the segment of a circle, see Geom- 
etry — Mensuration of Superficies. 



TONNAGE. 69 

TONNAGE. 

GOVERNMENT MEASUREMENT.* 



length — -I breadth X breadth X depth 

S J ~ = tonnage- 

ln a double-decked vessel, the length is reckoned from the fore 
part of the main stem to the after side of the sternpost above the 
upper deck ; the breadth is taken at the broadest part above the 
main wales, and half this breadth is taken for the depth. 

In a single-decked vessel the length and breadth are taken as for 
a double-decked vessel, and the distance between the ceiling of the 
hold and the under side of the deck plank is taken as the depth. 

Example. — The length of a double-decked vessel is 260 feet, and 
the breadth is 60 feet ; required the tonnage. 

260 ^-sP- = 224 X 60 X - 6 <f = 403200 -j- 95 = 4244.2 tons. Ans. 

Example. — The length of a single-decked vessel is 180 feet, the 
breadth 34 feet, and depth 18 feet ; required the tonnage. 

180 — f of 34 = 159.6 X 34 X 18 -r 95 = 1028.16 tons. Ans. 

CARPENTER'S MEASUREMENT. 

For a double-decked — 

length of keel X breadth main beam X h breadth 
g£ = tonnage. 

For a single-decked — 

length of keel X breadth main beam X depth of hold 

— 2 — = tonnage 

* The set of rules legalized by the Congress of 1865 for ascertaining the nomi- 
nal or Government tonnage of vessels are not inserted in this work, partly be- 
cause of their great lengths in detail, and the multiplicity of the mechanical ad- 
measurements required, and partly because they can be of use to but a few 
individuals; and to those they are furnished by the Government. 



70 CONDUITS OR PIPES. 

OF CONDUITS OR PIPES. 

Pressure of Water in Vertical Pipes, <5fC. 

h = height of column in inches ; o = circumference of column in inches; 
t — thickness of pipe in inches equal in strength to lateral pressure at base 
of column ; w = weight of a cubic inch of water in pounds- ; C = cohesive 
strength in pounds per inch area of transverse section of the material of 
which the pipe is composed — table, p. 74. 

ho = area of interior of pipe in inches ; hw = pressure in pounds per 
square inch at the base of the column, or maximum lateral pressure in 
pounds per square inch on the pipe tending to burst it ; how = maximum 
lateral pressure in pounds on the pipe, tending to burst it at the bottom ; 
and how ~ 2 = mean lateral pressure in pounds on the pipe, or pressure 
in pounds on the pipe tending to burst it at half the height of the column. 

how -j- C = t ; how -7- t = ; Gt -i- ow = h; Ct — hw = 0. 

Note. — The reliable cohesion of a material is not above I its ultimate force, as given 
in the Table of Cohesive Forces. By experiment, it has been found that a cast iron pipe 
15 inches in diameter and % of an inch thick, will support a head of water of 600 feet ; and 
that one of the same diameter made of oak, and two inches thick, will support a head of 
180 feet : 12000 lbs. per square inch for cast iron, 1200 for oak, 750 for lead, are counted 
safe estimates. The ultimate cohesion of an alloy, composed of lead 8 parts and zinc 1 
part, is 3000 pounds per square inch. 

Concerning the Discharge of Pipes, <Sfc. 

Small pipes, whether vertical, horizontal, or inclined, under equal 
heads, discharge proportionally less water than large ones. That 
form of pipe, therefore, which presents the least perimeter to its area, 
other things being equal, will give the greatest discharge. A round 
pipe, consequently, will discharge more water in a given time than a 
pipe of any other form, of equal area. 

The greater the length of a pipe discharging vertically, the greater 
the discharge. Because the friction of the particles against its sides, 
and consequent retardation, is more than overcome by the gravity of 
the fluid. 

The greater the length of a pipe discharging horizontally, the less 
proportionally will be the discharge. The proportion compared with 
a less length is in the inverse ratio of the square root of the two 
lengths, nearly. 

Other things being equal, rectilinear pipes give a greater discharge 
than curvilinear, and curvilinear greater than angular. The head, 
the diameters and the lengths being the same, the time occupied in 
passing an equal . quantity of water through a straight pipe is 9, 
through one curved to a semicircle 10, and through one having one 
right angle, otherwise straight, 14. All interior inequalities and 
roughness should be avoided. 

It has been ascertained that a velocity of 60 feet a minute ( 1 foot 
a second) through a horizontal pipe, 4 inches in diameter and 100 feet 



CONDUITS OR PIPES. 71 

in length, is produced by a head 2f inches, only f of an inch above 
the upper surface of the orifice ; and that, to maintain an equal 
velocity through a pipe similarly situated, of equal length, having a 
diameter of \ inch only, a head of ly 5 ^ feet is required. To increase 
the velocity through the last mentioned pipe to 2 feet a second, 
requires a head 4 T § feet ; to 3 feet, a head of lOy 1 ^- ; to 4 feet, a 
head of 17 T £, &c. 

From the foregoing, the following, it is believed, reliable rules, are 
deduced. 

To find the velocity of water passing through a straight horizontal pipe 
of any length and diameter, the head, or height of the fluid above the 
centre of the orifice, being known. 

Rule. — Multiply the head, in feet, by 2500, and divide the product 
by the length of the pipe, in feet, multiplied by 13.9, divided by the 
interior diameter of the pipe in inches ; the square root of the quotient 
will be the velocity in feet per second. 

Example. — The head is 6 feet, the length of the pipe 1340 feet, 
and its diameter 5 inches ; required the velocity of the water passing, 
through it. 

2500 X 6 = 15000 -T- (iMoxi3-9) __ ^4.03 — = 2 feet per 
second. Ans. 

To find the head necessary to produce a required velocity through a pipe 
of given length and diameter. 
Rule. — Multiply the square of the required velocity, in feet, per 
second, by the length of the pipe multiplied by the quotient obtained 
by dividing 13.9 by the diameter of the pipe in inches, and divide the 
product thus obtained by 2500 ; the quotient will be the head in feet. 

Example. — The length of a pipe lying horizontal and straight is 
1340 feet, and its diameter is 5 inches ; what head is necessary to 
cause the water to flow through it at the rate of 2 feet a second ? 
2 2 X 1340 X --P- 4- 2500 = 6 feet. Ans. 

To find the quantity of water flowing through a pipe of any length and 
diameter. 
Rule. — Multiply the velocity in feet per second by the area of the 
discharging orifice, in feet, and the product is the quantity in cubic 
feet discharged per second. 

Example. — The velocity is 2 feet a second, and the diameter of 
the pipe 5 inches ; what quantity of water is discharged in each 
second of time 1 
5 -h 12 = .4166, and .4166 2 X -7854 X 2 = .273 cubic foot. Ans. 



72 MISCELLANEOUS PROBLEMS. 

MISCELLANEOUS PROBLEMS. 

To find the specific gravity of a body heavier than water. 

Rule. — Weigh the body in water and out of water, and divide the 
weight out of water by the difference of the two weights. 

Example. — A piece of metal weighs 10 lbs. in atmosphere, and 
but 83 in water ; required its specific gravity. 

10 — 8.25 = 1.75, and 10 -H 1.75 = 5.714. Ans. 

To find the specific gravity of a body lighter than water. 
Rule. — Weigh the body in air ; then connect it with a piece of 
metal whose weight, both in and out of water, is known, and of suf- 
ficient weight that the two will sink in water ; and find their combined 
weight in water ; then divide the weight of the body in air by the 
weight of the two substances in air, less the sum of the difference of 
the weight of the metal in air and water and the combined weight of 
the two substances in water, and the quotient will be the specific 
gravity sought. 

Example. — The combined weight, in water, of a piece of wood, 
and piece of metal, is 4 lbs. ; the wood weighs in atmosphere 10 lbs. ; 
and the metal in atmosphere 12, and in water 11 lbs. ; required the 
specific gravity of the wood. 

10 -T- (10 -j- 12 — 12 ^11 + 4) = .588. Ans. 
To find the specific gravity of a fluid. 
Rule. — Multiply the known specific gravity of a body by the dif- 
ference of its weight in and out of the fluid, and divide the product by 
its weight out of the fluid ; the quotient will be the specific gravity of 
the fluid in which the body is weighed. 

Example. — The specific gravity of a brass ball is 8.6 ; its weight 
in atmosphere is 8 oz., and in a certain fluid 7^ oz. ; required the 
specific gravity of the fluid. 
8 — 7.25 = .75, and 8.6 X -75 = 6.45, and 6.45 -j- 8 = .806. Ans. 

To find the proportion of one to the other of two simples forming a 
compound, or the extent to which a metal is debased, (the metal and the 
alloy used being knoion.) 

The Rule strictly bears upon that of Alligation Alternate, which 
see. 

Example. — The specific gravity of gold is .19.258, and that of 
copper, 8.788; an article composed of the two metals, has a specific 
gravity of 18 ; in what proportion are the metals mixed 1 
18 ^ 19.258 X 8.788=11.055 
18 ^ 8.788 X 19.258 = 177.4, then 



MISCELLANEOUS PROBLEMS. 73 



11.055 + 177.4 : 11.055 : : 18 = 1.056 copper, > ^ 
11.055 -j- 177.4 : 177.4 : : 18 = 16.944 gold. S 
Or, 18 — 1.056 = 16.944 gold. Copper to gold as 1 to 16.04 + 

To find the lifting power of a balloon. 
Rule. — Multiply the capacity of the balloon, in feet, by the dif- 
ference of weight between a cubic foot of atmosphere and a cubic foot 
of the gas used to innate the balloon, and the product is the weight 
the balloon will raise. 

Example. — A balloon, whose diameter is 24 feet, is inflated with 
hydrogen ; what weight will it raise 1 . 

Specific gravity of air is 1, weight of a cubic foot 527.04 grains ; 
specific gravity of hydrogen is .0689. 

527.04 X - 0689 = 36 - 31 grains = weight of 1 cubic foot of hydrogen. 
527.04 — 36.31 = 490.73 grs. = dif. of weight of air and hydrogen. 
24 3 X -5236 = 7238.24 = capacity in cubic feet of balloon. 
Then, 7238.24 X 490.73 = 3552021 grs. = ^yW— = 507 T V lbs. 

• Ans. 

To find the diameter of a balloon that shall be equal to the raising of a 
given weight. 

The weight to be raised is 507^- lbs. 

507.4 X 7000-r-490.73 = 7238.21, and 7238.24 -f- .5236 = £/ 13824 
= 24 feet. Ans. 

To find the thickness of a concave or hollow metallic ball or globe, that shall 
have a given buoyancy in a given liquid. 

Example. — A concave globe is to be made of brass, specific grav- 
ity 8.6, and its diameter is to be 12 inches ; what must be its thick- 
ness that it may sink exactly to its centre in pure water? 

Weight of a cubic inch of water .036169 lb. ; of the brass .3112 lb. 
Then, 12 3 X -5236 X .036169 -i- 2 = 16.3625 cubic inches of water 
to be displaced. 

16.3625 -r- .3 112 = 52.5787 cubic inches of metal in the ball. 

12 2 X 3.1416 = 452.39 square inches of surface of the ball. 
And, 52.5787 -r- 452.39 = .1162 + = £ inch thick, full. Ans. 

To cut a square sheet of copper, tin, etc., so as to form a vessel of the 
greatest cubical capacity the sheet admits of. 

Rule. — From each corner of the sheet, at right angles to the side, 
cut ^ part of the length of the side, and turn up the sides till the 
corners meet. 

7 



74 



COMPARATIVE COHESIVE FORCE. 



Comparative Cohesive Force of Metals, Woods, and other substances, 
Wrought Iron (medium quality) being the unit of comparison, or 1 ; 
the cohesive force of which is 60000 lbs. per inch, transverse area. 



Wrought iron, . 


. 1.00 


Ash, white, ; 


. .23 


" " wire, . 


1.71 


" red, " . . 


. .30 


Copper, cast, . 


.40 


Beech, . • . 


.19 


" wire, . 


.76 


Birch, 


. .25 


Gold, cast, 


.34 


Box, 


.33 


. " wire, 


.51 


Cedar, 


.19 


Iron, cast, (average). 


.38 


Chestnut, sweet, • . 


.17 


Lead, " 


.015 


Cypress, . 


.10 


" milled, . 


.055 


Elm, 


.22 


Platinum, wire, 


.88 


Locust, . . 


.34 


Silver, cast, 


.66 


Mahogany, best, 


.36 


" wire, 


.68 


Maple, 


.18 


Steel, soft, 


2.00 


Oak, Amer., white, . 


. .19 


" fine, 


2.25 


Pine, pitch, 


.20 


Tin, cast block, 


.083 


Sycamore, 


.22 


Zinc, " ■ . . 


.043 


Walnut, . 


.30 


" sheet, 


.27 


Willow, . 


.22 


Brass, cast, 


.75 


Ivory, 


.27 


Gun metal, 


.50 


Whalebone, 


. .13 


Gold 5, copper 1, 


.83 


Marble, . 


.15 


Silver 5, " 1, . 


.80 


Glass, plate, 


.16 


Brick, 


.05 


Hemp fibres, glued, . 


. 1.53 


Slate, 


.20 







The strength of white oak to cast iron, is as 2 to 9. 
The stiffness of " " " " is as 1 to 13. 

To determine the weight, or force, in pounds, necessary to tear asun- 
der a bar, rod, or piece of any of the above named substances, of any 
given transverse area: 

Rule. — Multiply the comparative cohesive force of the substance, 
as given in the table, by the cohesive force per square inch, area of 
cross section (60000 lbs.) of wrought iron, which gives the cohesive 
force of 1 square inch area of cross section of the substance whose 
power is sought to be ascertained, and the product of 1 square inch 
thus found, multiplied by area of cross section, in inches, of the rod, 
piece, or bar itself, gives the cohesive force thereof. 

Alloys having a tenacity greater than the sum of their constituents. 
Swedish copper 6 pts., Malacca tin 1 ; tenacity per sq. inch, 64000 lbs. 
Chili copper 6 pts., Malacca tin 1; " " " 60000" 

Japan copper 5 pts., Bunca tin 1 ; " " " 57000 " 

Anglesea copper 6 pts., Cornish tin 1 ;. " " " 41000 " 



LINEAR DILATION OF SOLIDS BY HEAT. 



75 



Common block-tin 4 pts., lead 1, zinc 1 ; tenacity per sq. in., 13000 lbs. 
Malacca tin 4 pts., regulus of antimony 1 ; " " " 12000 " 

Block-tin 3 pts., lead 1 part ; ." " " 10000 " 

Block-tin 8 pts., zinc 1 part; " " " 10200" 

Zinc 1 part, lead 1 part ; " " " 4500 " 

Alloys having a density greater than the mean of their constituents. 

Gold with antimony, bismuth, cobalt, tin, or zinc. 

Silver with antimony, bismuth, lead, tin, or zinc. 

Copper with bismuth, palladium, tin, or zinc. 

Lead with antimony. 

Platinum with molybdinum. 

Palladium with bismuth. 

Alloys having a density less than the mean of their constituents. 
Gold with copper, iron, iridium, lead,, nickel, or silver. 
Silver with copper or cobalt. 
Iron with antimony, bismuth, or lead. 
Tin with antimony, lead, or palladium. 
Nickel with arsenic. 
Zinc with antimony. 

RELATIVE POWER OF DIFFERENT METALS TO CONDUCT ELEC- 
TRICITY, 
(the mass of each being equal.) 

Copper, .... 1000 1 Platinum, . . .188 

Gold, .... 936 1 Iron, .... 158 

Silver, .... 736 1 Tin, .... 155 

Zinc, . . ... 285 J Lead, . . . . 83 



LINEAR DILATION OF SOLIDS BY HEAT. 

Length which a bar heated to 212° has greater than when at the tem- 
perature of 32°. 



Brass, cast, 


. .0018671 


Iron, wrought, . 


.0012575 


Copper, 


. .0017674 


Lead, 


.0028568 


Fire brick, 


. .0004928 


Marble, 


.0011016 


Glass, 


. .0008545 


Platinum, . 


.0009342 


Gold, 


. .0014880 


Silver, 


.0020205 


Granite, . 


. .0007894 


Steel, 


.0011898 


Iron, cast, 


. .0011111 


Zinc, 


.0029420 



Note. —To find the surface dilation of any particular article, double its linear dilation, 
and to find the dilation in volume, triple it. To find the elongation in linear inches per 
linear foot, of an J particular article, multiply its respective linear dilation, as given in the 
1ABLK. by 12. 



76 



EFFECTS OF HEAT. 



MELTING POINT OF METALS AND OTHER BODIES. 

Lime, palladium, platinum, porcelain, rhodium, silex, may be melted 
by means of strong lenses, or by the hydro-oxygen blowpipe. Co- 
balt, manganese, plaster of Paris, pottery, iron, nickel, &c, at from 
2700° to 3250° Fahrenheit ; others as follows : — 





Degrees Fah. 








De 


sprees Fah. 


Antimony, . . 809 


Nitre, 


Beeswax, bleached, . .155 


Silver, 


. 1873 


Bismuth, .... 506 


Solder, common, 


. 475 


Brass, .... 1900 


" plumbers', 


. 360 


Copper, .... 1996 


Sugar, 


. 400 


Glass, flint, . . .1178 


Sulphur, 






. 226 


Gold, .... 2016 


Tallow, 






. 127 


Lead, . . . .612 


Tin, . 






. 442 


Mercury, .... — 39 


Zinc, 






. 680 


Cast iron thoroughly melts at ... 






■2786 


Greatest heat of a smith's forge, (com.) . 






2346 


Welding heat of iron, 






1892 


Iron red hot in twilight, 






884 


Lead 1, tin 1, bismuth 4, melts at . . 






. 201 


Lead 2, tin 3, bismuth 5, il " . 






. 212 


RELATIVE POWER OF DIFFERENT BODIES TO RADIATE HEAT. 


Water, .... 100 


Lead, bright, ... 19 


Copper, . 




12 


Mercury, 




20 


Glass, 




90 


Paper, white, . 




100 


Ice, 




85 


Silver, 




12 


India ink, 




88 


Tin, blackened, 




100 


Iron, polished, 




15 


" clean, 




12 


Lampblack, 




100 


" scraped, . 




16 


Note. — The power of a body to reflect heat is inverse to its power of radiation. 


BOILING POINT OF LIQUIDS. 


Barometer at 30 in. 


Acid, nitric, . . 253° 


Oils, essential, avg., . 318° 


" sulphuric, . . 600° 


" turpentine, . . 316° 


Alcohol, anhyd., . . 168.5° 


" linseed, . . 640° 


" 90 per cent., . 174° 


Phosphorus, . . 554° 


Ether, sulph., . . 97° 


Sulphur, . . . 560° 


Mercury, 




656° 


Water, 






. 


. 212° 



Note. — Barometer at 31 inches, water boils at 2130.57; at 29, it boils at21CP.38; at 
28, it boils at 2080.69 ; at 27, it boils at 2060.85, and in vacuo it boils at 88°. No liquid, 
under pressure of the atmosphere alone, can be heated above its boiling point. At that 
point the steam emitted sustains the weight of the atmosphere. 



EFFECTS OF HEAT, ETC. 



77 



FREEZING POINT OF LIQUIDS. 



Acid, nitric, 


. —55° 


Oil, linseed, avg., 


—11° 


" sulphuric, 


1° 


Proof spirits, 


—7° 


Ether, . 


— 47° 


Spirits turpentine, 


10° 


Mercury, 


. —39° 


Vinegar, 


28° 


Milk, . 


30° 


Water, 


32° 


Oil, cinnamon, 


30° 


Wine, strong-, 


20° 


" fennel, . 


14° 


Rapeseed Oil, 


25° 


" olive, 


36° 







EXPANSION OF FLUIDS BY BEING HEATED FROM 32° TO 212°, F. 



Atmospheric air, ^-^ per each degree, 
Gases, all kinds, ¥ ^ " " " 
Mercury, exposed, . 
Muriatic acid, (sp. gr. 1.137,) 
Nitric acid, (sp. gr. 1.40,) 
Sulphuric acid, (sp. gr. 1.85,) . 

" ether, — to its boiling point, 

Alcohol, (90 per cent.,) " " 

Oils, fixed, 

" turpentine, . . . . 
Water, 



= .375 



.018 
.060 
.110 
.060 
.070 
.110 
.080 
.070 
.046 



RELATIVE POWER OF SUBSTANCES TO CONDUCT HEAT. 



Gold, 
Silver, . 
Copper, . 
Platinum, 
Iron, 



1000 
973 
898 
381 
374 



Zinc, 
Tin, 
Lead, 
Porcelain, 
Fire brick, 



363 

304 

180 

12 

11 



Note. — Different wood3 have a conducting power in ratio to each other, as is their 
respective specific gravities, the more dense having the greater. 



METALS IN ORDER OF DUCTILITY AND MALLEABILITY. 



Ductility. 

1. Platinum. 

2. Gold. 

3. Silver. 

4. Iron. 

5. Copper. 

6. Zinc. 

7. Tin. 
8.- Lead. 



Malleability. 

1. Gold. 

2. Silver. 

3. Copper. 

4. Tin. 

5. Platinum. 

6. Lead. 

7. Zinc. 

8. Iron. 



7* 



78 



NUTRITIVE AND ALCOHOLIC PROPERTIES OF BODIES. 



Quantity per cent, by weight of Nutritious Matter contained in different 
articles of, Food. 



Articles. 






per ct. 


Articles. 






per ct. 


Lentils, . 


94 


Oats, .... 74 


Peas, 






93 


Meats, avg., 






35 


Beans, 






92 


Potatoes, . 






25 


Corn, (maize,) 






89 


Beets, 






14 


Wheat, . 






85 


Carrots, . 






10 


Barley, . 






83 


Cabbage, . 






7 


Rice, 






88 


Greens, . 






6 


Rye, 






79 Turnips, white, 




4 


Specific gravity, and quantity per cent., by volume, of Absolute Alcohol 


contained, necessary to constitute the following named unadulterated 


articles. 




Articles. 


Sp. £rrav. 
60°, b. 30. 


Per cent, 
of Alcohol. 


Absolute Alcohol, (anhydrous,) . . . .7939 


100 


Alcohol, highest by distill* 


ition, . . . .825 


92.6 


" commercial stand 


ard, . . . .8335 


90 


Proof Spirits, — standard, 


.9254 


54 


Quantity per fent., by volume, (general average) of Absolute Alcohol 


contained in different pu 


re or unadulterated Liquors, Wines, Sfc. 


Liquors, &c. 


per cent. 


Wines. per cent. 


Rum, 


50 


Port, . . -22 


Brandy, . 




50 


Madeira, 




20 


Gin, Holland, . 




4.8 


Sherry, 




18 


Whiskey, Scotch, 




50 


Lisbon, . 






17 


" Irish, 




50 


Claret, 






10 


Cider, whole, 




9 


Malaga, . 






16 


Ale, 




8 


Champagne, 






14 


Porter, 




7 


Burgundy, 






12 


Brown Stout, . 




6 


Muscat, 


* 




17 


Perry, 






9 


Currant, 






19 



Proof of Spirituous Liquors. 

The weight, in air, of a cubic inch of Proof Spirits, at 60° F., is 
233 grains ; therefore, an inch cube of any heavy body, at that tempera- 
ture, weighing 233 grains less in spirits than in air, shows the spirits 
in which it is weighed to be proof. If the body lose less of its weight, 
the spirit is above proof, — if more, it is below. 



COMPARATIVE WEIGHT OF TIMBER. 



79 



Comparative Weight of different kinds of Timber in a green and per- 
fectly seasoned state. 
Assuming- the weight of each kind destitute of water to be 100, that 
of the same kind green is as follows : — 



Ash, 


153 1 Cedar, . 


148 1 Maple, red, . 


149 


Beech, . 


174 j Elm, swamp. . 


198 1 Oak, Am., . 


151 


Birch, . 


169 1 Fir, Amer., '. 


171 1 Pine, white, . 


152 



Note. — Woods which have been felled, cleft and housed for 12 months, still retain 
from 20 to 25 per cent, of water. They therefore contain but from 75 to 80 per cent, of 
heating matter : and it will require from 23 to 29 per cent, the weight of such woods to 
dispel the water they contain. They are, therefore, less valuable by weight, as fuel, by 
this per cent., than woods perfectly free from moisture. They never, however, contain, 
exposed to an ordinary atmosphere, less than 10 per cent, of water, however long kept; 
and even though rendered anhydrous by a strong heat, they again imbibe, on exposure to 
the atmosphere, from 10 to 12 per cent, of dampness. 

Relative poioer of different seasoned Woods, Coals, 6fC.,asfuel, to pro- 
duce heat, — the Woods supposed to be seasoned to mean dryness, 
(77£ per cent.,) and the other articles to contain but their usual quan- 
tity of moisture. 





Ratio of Heating 
Power per equal 




Bulk. 


Weight. 


Hickory, shell-bark, 


1.00 


1.00 


" red-heart, 














.81 


.99 


Walnut, com. 














.95 


.98 


Beech, red, 














.74 


.99 


Chestnut, 














.49 


.98 


Elm, white, 














.58 


.98 


Maple, hard, 














.66 


.98 


Oak, white, 














.81 


.99 


" red, 














.69 


.99 


Pine, white, 














.42 


1.01 


" yellow, 














.48 


1.03 


Birch, black, 














.63 


.99 


" white, 














.48 


.99 


Coal, Cumberland, (bit.) 












2.56 


2.28 


" Lackawanna (anth.) 












2.28 


2.22 


" Lehigh, " 












2.39 


2.03 


" Newcastle, (bit.) 












2.10 


1.96 


" Pictou, (bit.) 












2.21 


1.91 


" Pittsburgh, (bit.) 












1.78 


1.82 


" Peach Mountain, (anth. 


) 










2.69 


2.29 


Charcoal, .... 












1.14 


2.53 


Coke, Virginia, natural 














1.89 


2.J2 


" Cumberland, 














1.31 


2.25 


Peat, ordinary, . 
















62 


Alcohol, common, 
















2.02 


Beeswax, yellow, 
















2.90 


Tallow, . 
















3 10 



80 



ILLUMINATION. 



Note. — By help ot the preceding table, the price of either one article being; known, 
the relative or par value of either other, as fuel, may be readily ascertained : — Example: 

Maple (66) : $5.00 : : Pine (42) : $3.18. 



ILLUMINATION — ARTIFICIAL. 

The following Table shows : — 

1 . The materials and methods of using - — column Materials. 

2. The comparative maximum intensity of light afforded by each 
material, used or consumed as indicated, — column Intensities. 

3. The weight, in grains, of material consumed per hour, by each 
method respectively, in producing its respective light, or light of in- 
tensity ascribed — column Weight. 

4. The ratio of weight required of each material, under each spe- 
cial method of consumption, for the production of equal lights in 
equal times — column Ratios. 

Materials. 

Camphene, Paragon Lamp, 

Sperm Oil, Parker's heating Lamp, 

" " Mech. or Carcel " 

" " French annular " 

" " Common hand " 

Whale " p'f'd., P's heating " 
Wax Candles, 3's or 4's, 15 in. or 12 in., 

" " 6's, 9 in., 
Sperm " 4's, 13£ in., . 
Stearine 11 4's, 13£ in., . 
Tallow, " dipped, 10's, . 

" " mould, 10's, . 

" " " 8's, . 

" " " 6's, 

4's, 131 in., 
" Coal Gas," intensity being 

Note. — The consumption of 1.43 cubic feet of gas per hour, gives a light equal to one 
wax candle, — the consumption of 1.96 cubic feet per hour, a light equal to four wax can- 
dles, and the consumption of 3 cubic feet per hour, a light equal to ten wax candles. A 
cubic fool of gas weighs 286 grains. 

The average yield of bi-carbureted hydrogen — defiant gas — Coal Gas. obtained from 
the following articles, is as annexed. 

1 lb. Bituminous Coal, . . . . ■ . 

1 lb. Oil, or Oleine, 

1 lb. Tar 

1 lb. Rosin, or Pitch, 

A pound of good Lancashire cannel coal, or of good 
yield, on an average, 5.95 cubic feet of good illuminating gas. 

From the English Boghead cannel, by "White's hydro-carbon process, 17 cubic 
feet to the pound are commonly obtained. 

A pipe whose interior diameter is £ inch, will supply gas equal in illuminating 
power to 20 wax candles. 

The sp. gr. of coal-gas from cannel coal ranges from 0.640 to 0.370, while that 
from good working bituminous coal in general ranges from 0.420 to 0.370. 





Intens. 


Weia-ht. 


Ratios. 




16. 


853 


1. 




11. 


696 


1.19 




10. 


815 


1.53 




5. 


543 


2.04 




L 


112 


2.10 




9. 


780 


1.63 




1. 


125 


2.35 




.92 


122 


2.50 




1. 


142 


2.66 




1. 


168 


3.15 




.70 


150 


4.02 




.66 


132 


3.75 




.57 


132 


4.35 




.79 


163 


3.87 




I. 


186 


3.49 




1. 


411 





4^ cubic feet. 
15 " 
12 " 
10 " 
Scotch cannel, will 



ILLUMINATION. 



81 



Results of Experiments by Mr. Clegg of London, relative to the 
conveyance of coal-gas through pipes of different lengths, diame- 
ters, fyc. 



Diameter of pipe 
in inches. 


Length of pipe 
in yards. 


Pressure in inches 
of water. 


Specific gravity, 
air as 1. 


Quantity in cubic 

feet discharged 

per hour. 


0.5 


10 


1.25 


0.4 


120 


0.5 


59 


1.25 


0.4 


60 


0.62 


41 


1.38 


0.559 


99 


0.62 


62 


1.34 


0.559 


83 


0.62 


93 


1.34 


0.559 


74 


0.62 


119 


1.34 


0.559 


57 


0.62 


138 


1.34 


0.559 


53 


2. 


25 


0.5 


0.528 


1630 


10. 


100 


3. 


.0.4 


120,000 


10. 


1760 


3. 


0.4 


30,000 


18. 


1760 


1. 


0.4 


66,000 


26. 


4300 


0.475 


0.42 


80,000 


26. 


3130 


0.8 


0.42 


103,000 


26. 


4300 


2.25 


0.42 


175,000 



From these experiments and others, Mr. Clegg derives the data 
for the following general approximate rules, viz. : 

Q = 1350d 2 ^j ; or, when the length of the pipe is above 

i ol L I ■ Li ) 

400 or 500 times its diameter ; or, in any case, if the pressure be 
measured in the pipe instead of in the gas-holder, Q = 1350c? 2 5 

in which Q represents the quantity of gas in cubic feet passed per 
hour, I the length of the pipe in yards, d the diameter of the pipe 
in inches, h the pressure in inches of water, and s the specific 
gravity of the gas. 

Therefore, ordinarily for short mains of large diameters, or 
when the quotient of 36/-^- d is less than 400, 



1350~c7 5 A _ d . h = Qi s (l+d) . 



and I -f- d = 



1350 2 d 5 h 



Q-s 1350 2 6? 5 ' ' Q-s 

And for long pipes of small diameters, or when I and I -\- d are 
practically equal. 



1 = 



1350'd=h 



_ Oris 



and d 



_ 5/ Q 2 ls 



4 



Q 2 s 1350-d 5 ' ^1350-'A 

Now, in these formulas for Q, by Mr. Clegg, the probable re- 
tardation of the flow of the gas due to the friction of its particles 
in the pipes has been taken into account ; and the results by the 
formulas agree as closely as could be expected with the given ex- 
periments, although the former average about 7 per cent, more 
than the latter. 



82 THERMOMETERS. 

The pipes are here supposed to be straight, and to lie horizon- 
tal, or equal in effect to that condition, and the gas is supposed to 
be delivered without force at the discharging end. 

Example. — What pressure in inches of water is required to 
convey 852 cubic feet of gas per hour, of specific gravity 0.398, 
through a pipe 4 inches in diameter and 6 miles in length V 

852 2 X 10560 X =398 , OOAO . ', A 

Ci a =r 1.6348 inches. Ans. 

1350 2 X 4 5 

Note. — The illuminating power of coal-gas is nearly as its "specific gravity, 
the more dense being the better. 



THERMOMETERS. 

>ihng point. 



Fahrenheit's, 

Reaumur's, 

Centigrade, 



212^ 

80 c 

100 c 



Freezing point. 

32 s 
0° 

0° 



To reduce Reaumur to Fahrenheit. 
When it is desired to reduce the -\-°, (degrees above the zero) : — 

Rule. — Multiply the degrees Reaumur, by 2.25, and add 32° to 
the product ; the sum will be the degrees Fahrenheit. 

When it is desired to reduce the — °, (degrees below the zero) : — 

Rule. — Multiply the — ° Reaumur by 2.25, and subtract the 
product from 32° ; the difference will be the degrees Fahrenheit. 

Example. — The degrees R. are 40 ; required the equivalent 
degrees F. 

40 X 2.25 = 90 -j- 32 = 122°. Ans. 

Example. — The degrees below 0, R., are 10 ; what are the cor- 
responding degrees F. ? 

10 X 2.25 = 22.5, and 32 — 22.5 = 9£°. Ans. 

Example. — The degrees below 0, R., are 16 ; what point on the 
scale F. corresponds thereto 1 

16 X 2.25 = 36, and 32 — 36 = — 4 : 4° below 0. Ans. 

To reduce the Centigrade to Fahrenheit. 
Rule. — Multiply the degrees C. by 1.8, and in all other respects 
proceed as directed for Reaumur, above. 

Note. — The zero of Wedgewood's pyrometer is fixed at the temperature of iron red-hot 
in daylight, = 1077° F., and each degree W. equals 130° F. The instrument is not con- 
sidered reliable, and is but little used 



HORSE POWER ANIMAL POWER STEAM. 83 

HORSE POWER. 

A horse-power, in machinery, as a measure of force, is estimated 
equal to the raising of 33000 lbs. over a single pulley one foot a 
minute, = 550 lbs. raised one foot a second, = 1000 lbs. raised 33 feet 
a minute. 



ANIMAL POWER. 

A man of ordinary strength is supposed capable of exerting a force 
of 30 lbs. for 10 hours in a day, at a velocity of 2£ feet a second, = 
75 lbs. raised 1 foot a second. 

The ordinary working power of a horse is calculated at 750 lbs. for 
8 hours in a day, at a velocity of 2 feet a second, = 375 lbs. raised 
1 foot a second, = 5 times the effective power of a man during asso- 
ciated labor, and 4 times his power per day ; and as machinery may 
be supposed to work continually, = a trifle less than 23 per cent, per 
day of a machine horse-power. 



STEAM. 

Table exhibiting the expansive force and various conditions of steam 
under different degrees of temperature. 



Degrees of 


Pressure in 


Density. 


Volume. 


Spec, gravity. 


Weight of a 


heal. 


atmospheres. 


Water as 1. 


Water as 1. 


Air as 1. 


grains. 


212 


1 


.00059 


1694 


.484 


254 


250.5 


2 


.00110 


909 


.915 


483 


270 


3 


.00160 


625 


1.330 


700 


293.8 


4 


.00210 


476 


1.728 


910 


308 


5 


.00258 


387 


2.120 


1110 


359 


10 


.00492 


203 


3.970 


2100 


418.5 


20 


.00973 


106 


7.440 


3940 



[An atmosphere is 14-^ lbs. to the square inch.] 

Note. — By the above table it is seen that any given quantity of steam having a tem- 
perature of 212° F., occupies a space, under the ordinary pressure of the atmosphere, 
169-1 times greater than it occupied when as water in a natural state. It exerts a mechan- 
ical force, consequently, = 1694 times the weight or force of the atmosphere resting on 
the bulk from which it was generated, or resting on l-1694th of the space it occupies. 
A force, if we consider the volume as so many cubic inches, equal to the raising of 2087 
lbs. 12 inches high, by a quantity of steam less than a cubic foot, heated only to the tem- 
perature of boiling water, and.weighing but 248 grains, and that, too, the product of a 
single cubic inch of water. 

The mean pressure of the atmosphere at the earth's surface is equal 
to the weight of a column of mercury 29.9 inches in height, or to a 
column of water 33.87 feet in height, = 2116.8 lbs. per square foot, or 



84 



VELOCITY AND FORCE OF WIND. 



14.7 lbs. per square inch. Its density above the earth is uniformly 
less as its altitude is greater, and its extent is not above 50 miles — ■ 
its mean altitude is about 45 miles ; at 44 miles it ceases to reflect 
light. Were it of uniform density throughout, and of that at the sur- 
face, its altitude would be but 5| miles. Its weight is to pure water 
of equal temperature and volume, as 1 to 829. It revolves with the 
earth, and its average humidity, at 40° of latitude, is 4 grains per cubic 
foot. Its weight at 60°, b. 30, compared with an equal bulk of pure 
water at 40°, b. 30, is as 1 to 830.1. 



VELOCITY AND FORCE OF WIND. 





Mean ve 


locily in 


1 




Miles per 


Feet per 


Force in lbs. per 


Appellations. 


hour. 


second. 


square foot. 


Just perceptible, 


2£ 


3| 


.032 


Gentle, pleasant wind, . 


4£ 


6| 


.101 


Pleasant, brisk gale, . 


m 


18£ 


.80 


Very brisk, " 


22h 


33 


2.52 


High wind, 


32h 


47| 


5.23 


Very high wind, . 


42£ 


62£ 


8.92 


Storm, or tempest, 


50 


73i 


12.30 


Great storm, 


60 


88 


17.71 


Hurricane, 


80 


117| 


31.49 


Tornado, moving buildings, &c, 


100 


146.7 


49.20 



The curvature of the earth is 6.99 inches (.5825 foot) in a single 

statute mile, or 8.05 inches in a geographical mile, and is as the 

square of the distance for any distance greater or less, or space 

between two levels ; thus, for three statute miles it is 

1 : 3 2 : : 6.99 : 5| feet, nearly. 

The horizontal refraction is y^. 

Degrees of longitude are to each other in length, as the cosines of 
their latitudes. At the equator a degree of longitude is 60 geographical 
miles in length, at 90° of latitude it is ; consequently, a degree of 
longitude at 



5° 


r= 59.77 miles. 


40° 


= 45.96 miles. 


10° 


= 59.09 " 


50° 


= 38.57 " 


20° 


= 56.38 " 


70° . 


= 20.52 " 


30° 


= 51.96 " 


85° . 


= 5.23 " 



Time is to longitude 4 minutes to a degree, — faster, east of any 
given point ; slower, west. 

The mean velocity of sound at the temperature of 33° is 1100 feet 
a second. Its velocity is increased £ a foot a second for every degree 



GRAVITATION. 85 

above 33°, and decreased \ a foot a second for every degree below 
33 . 

In water, sound passes at the rate of 4,708 feet a second. 

Light travels at the rate of 192,000 miles per second. 

GRAVITATION. 

Gravity, or Gravitation, is a property of all bodies, by which 
they mutually attract each other proportiocally to their masses, 
and inversely as the square of the distance of their centres apart. 
Practically, therefore, with reference to our Earth and the bodies 
upon or near its surface, gravity is a constant force centred at the 
Earth's centre, and is there continually operating to draw all bodies 
with a uniformly accelerating velocity to that point, and through 
very nearly equal spaces, in equal intervals of time from rest, at all 
localities. 

Putting R< to represent the Equatorial radius of the earth, and r 
to represent the Polar, and making R! = 3962.5 statute miles, and 
rz=z 3949.5, which is nearly in accordance with the mean of the 
most reliable measurements of the arcs of a degree of latitude at 
different localities, we have e 2 =(R' 2 — r 2 ) ~R> 2 = .006550751, the 
square of the ellipticity of the earth, and R = 2R' — (2 -4- e 2 sin 2 Z^), 
the radius at any given latitude I. 

And since the initial velocity due to gravity at the level of the 
sea at the Equator is (7 = 32.0741 feet per second, or, in other 
words, since a body falling in vacuo at the equator, at the level of 
the sea, describes a space of 16.03705 feet in the first second of 
time from rest, we have g=i\_R' y'G -f- .R] 2 , the initial velocity at 
the level of the sea at any given radius R ; or g = (22441.2 — R) 2 . 

And finally ,, = (!?±^y x (l-^) at any given ra- 

dius R, at any given altitude, h, in feet, above the level of the sea. 

Note. — When I, reckoned from the equator, is higher than 45°, sin 2 I = 
cos2(90 — I). 

The momentum, or force, with which a falling body strikes, is the 
product of its weight and velocity (the weight multiplied by the 
square root of the product of the space fallen through and 64.33, 
or 4 times 16-^) ; thus, 100 lbs., falling 50 feet, will strike with a 
force, 

50 X 64.333 = ^/ 3216.66 = 56.71 X 100 = 5671 lbs. 

An entire revolution of the earth, from west to east, is performed 
in 23 hours, 56 minutes, and 4 seconds. A solar year = 365 days, 
5 hours, 48 minutes, 57 seconds. 

The area of the earth is nearly 1 9 7,000,000 square miles. Its crust 

is supposed to be about 30 miles in thickness, and its mean density 5 

times that of water. About -f of its area, or 150,000,000 square 

miles, is covered by water. The portions of land in the several 

8 



86 



CHEMICAL ELEMENTS. 



divisions, in square miles, are, in round numbers, as follows, 
viz: — 

Asia, . . 16,300,000 1 Europe, . . 3,700,000 

Africa, . . 11,000,000 Australia, . . 3,000,000 

America, . . 14,500,0001 

America is 9000 miles long-, or y 3 ^ the circumference of the 
earth. 

The population of the globe is about 1,000,000,000, of which there 
are, in 

Asia, . . 456,000,000 I Africa, . . 62,000,000 

Europe, . . 258,000,000 | America, . . 55,000,000 



CHEMICAL ELEMENTS. 

The chemical elements — simple substances in nature — as far as 
has been determined, are 62 in number : 13 non-metallic and 49 
metallic. 

Of the non-metallic, 5 — bromine, chlorine, fluorine, iodine, and oxy- 
gen, (formerly termed "supporters of combustion") have an intense 
affinity for all the others, which they penetrate, corrode, and appar- 
ently consume, always with the production, to some extent, of light 
and heat. They are all non-conductors of electricity and negative 
electrics. 

The remaining 8 — hydrogen, nitrogen or azote, carbon, boron, sili- 
con, phosphorus, selenium, and sulphur, are eminently susceptible of 
the impressions of the preceding five ; when acted upon by either of 
them to a certain extent, light and heat are manifestly evolved, and 
they are thereby converted into incombustible compounds. 

Of the metals, 7 — potassium, sodium, calcium, barytium, lithium, 
strontium, and magnesia, by the action of oxygen, are converted into 
bodies possessed of alkaline properties. 

Seven of them — glucinum, erbium, terbium, yttrium, allumium, zir- 
conium, and thorium, — by the action of oxygen, are converted into 
the earths proper. 

In short, all the metals are acted upon by oxygen, as also by most 
or all of the non-metallic family. The compounds thus formed are 
alkaline, saline, or acidulous, or an alkali, a salt, or an acid, according 
to the nature of the materials and the extent of combination. 

Metals combine with each other, forming alloys. If one of the 
metals in combination is mercury, the compound is called an amalgam. 

Silicon is the base of the mineral world, and carbon of the organ- 
ized. 

For a very general list of the metals, see Table of Specific Grav- 
ities. 



CONSTITUENTS OF BODIES. 



87 



TABLE 

Exhibiting the Elementary Constituents and -per cent, by weight of each, 
in 100 parts of different compounds. 



Compounds. 

Atmospheric air, . 

Water, pure, 

Alcohol, anhydrous, 

Olive oil, . 

Sperm" 

Castor " 

Stearine, (solid of fats,) 

Oleine, (liquid of fats,) 

Linseed oil, . 

Oil of turpentine, 

" Camphene" (pure spts. turp 

Caoutchouc, (gum elastic,) 

Camphor, 

Copal, resin, 

Guaiac, resin, 

Wax, yellow, 

Coals, cannel, 

" Cumberland, 

" Anthracite, 
Charcoal, 
Diamond, 
Oak wood, dry, . 
Beech " " 
Acetic acid, dry, 
Citric " crystals, 
Oxalic " dry, 
Malic, " crystals, 
Tartaric " dry, 
Formic " " 
Tannin, tannic acid, solid 
Nitric acid, dry, . 
Nitrous " anhydrous, liquid 
Ammoniacal gas, 
Carbonic acid " . 
Carb. hydrogen gas, 
Bi-carb. hyd., olefient gas, 
Cyanogen " 

Nitric oxyde " 

Nitrous " " 

Ether, sulphuric, . 
Creosote, 



Constituents 


md per cent 




Hydrogen. 


Oxygen. 


Nitrozen. 


Carbon. 




19.96 


79.84 




11.1 


88.9 






12.9 


34.44 




52.66 


13.4 


9.4 




77.2 


10.97 


10.13 




78.9 


10.3 


15.7 




74.00 


11.23 


6.3 


0.30 


82.17 


11.54 


12.07 


0.35 


76.03 


11.35 


12.64 




76.01 


11.74 


3.66 




84.6 


11.5 






88.5 


10. 






90. 


11.14 


11.48 




77.38 


9. 


11.1 




79.9 


7.05 


25.07 




67.88 


11.37 


7.94 




80.69 


3.93 


21.05 


2.80 


72.22 


3.02 


14.42 


2.56 


80. 
93. 
97. 
100. 


5.69 


41.78 




52.53 


5.82 


42.73 




51.45 


5.82 


46.64 




47.54 


4.5 


59.7 




35.8 




79.67 




20.33 


3.51 


55.02 




41.47 


3. 


60.2 




36.80 


2.68 


64.78 




32.54 


4.20 


44.24 




51.56 




73.85 


26.15 






61 32 


30.68 




17.47 




82.53 






72.32 




27.68 


24.51 






75.49 


14.05 






85.95 






53.8 


46.2 




53. 


47.00 






36.36 


63.64 




13.85 


21.24 




65.05 j 


7.8 


16. 




76.2 ! 



CONSTITUENTS OF BODIES. 





Constituents 


and per cent. 


Compounds. 

Morphia, .... 


Hydrogen. 


Oxygen. 


Azote. 


Carbon. 

72.34 


6.37 


16.29 


5. 


Quina, — quinine, 




7.52 


8.61 


8.11 


75.76 


Veratrine, .... 




8.55 


19.61 


5.05 


66.79 


Indigo, 






4.38 


14.25 


10. 


71.37 


Silk, pure white, 






3.94 


34.04 


11.33 


50.69 


Starch, — farina, < 


iextrine, 




6.8 


49.7 




43.5 


Sugar, 






6.29 


50.33 




43.38 


Gluten, 






7.8 


22. 


14.5 


55.7 


Wheat, 




c 


6. 


44.4 


2.4 


47.02 


Rye, . 






5.7 


45.3 


1.7 


47.03 


Oats, . 






6.6 


38.2 


2.3 


52.9 


Potatoes, 






6.1 


46.4 


1.06 


45.9 


Peas, . 






6.4 


41.3 


4.3 


48. 


Beet root, 






6.2 


46.3 


1.8 


45.7 


Turnips, 






6. 


45.9 


1.8 


46.3 


Fibrin, 




d 


7.03 


20.30 


19.31 


53.36 


Gelatin, 




d 


7.91 


27.21 


17. 


47.88 


Albumen, 




d 


7.54 : 23.88 


1 15.70 ' 52.88 



Muriatic acid gas, — Hydrogen 5.53 -\- 94.47 chlorine. 
Sulphuric acid, dry, — Oxygen 79.67 -|- 20.33 sulphur. 
Silicic acid — Silica, dry, — Oxygen 51.96 -\- 48.04 silicon. 
Boracic acid — Borax, dry, — " 68.81 -|- 31.19 boron. 

a. The atmosphere, in addition to its constituents as given in the 
table, contains, besides a small quantity of vapor, from 1 to 3 parts in 
a thousand of carbonic acid gas, and a trace merely of ammoniacal gas. 

b. Anthracite coal, charcoal, plumbago, coke, &c, have no other 
constituent than carbon ; they are combined, to a small extent, with 
foreign matters, such as iron, silica, sulphur, alumina, &c. 

c. The constituents of woods, grains, &c, are given per cent., with- 
out regard to the foreign matters {metallic) which they contain. In 
oak, chestnut, and Norway pine, the ashes amount to about -^ of 1 per 
cent., and in ash and maple to JL. of 1. In anthracite coals, at an 
average, they are about 7 per cent. 

d. Fibrin, Gelatin, Albumen — Proximate animal constituents — 
Nutritious properties of animal matter. 

Fibrin is the basis of the muscle (lean meat) of all animals, and is 
also a large constituent of the blood. 

Gelatin exists largely in the skin, cartilages, ligaments, tendons and 
bones of animals. It also exists in the muscles and the membranes. 

Albumen exists in the skin, glands and vessels, and in the serum of 
the blood. It constitutes nearly the whole of the white of an egg. 



CONSTITUENTS OV BODIES. £9 

The relative quantities by volume of the several gases going to 
constitute any particular compound, are readily ascertained by help 
of their respective specific gravities, compared with their relative 
weights, as given per cent, in the preceding table: — thus, the sp. 
gr. of hydrogen is .0689, and that of oxygen 1.1025, and 1.1025 -7- 
.0689 = 16 ; showing the weight of the latter to be 16 times that of 
the former per equal volumes, or, relatively, as 16 to 1. The per 
cent, by weight, as shown by the table, in which these two gases 
combine to form water, for instance, is 11.1 and 88.9 ; or 11.1 of 
hydrogen and 88.9 of oxygen in 100 of the compound ; or as 88.9 -j- 
11.1, — as 8 to 1 : 16 -7- 8 = 2 : two volumes, therefore, of the 
lighter gas (hydrogen) combine with one of oxygen to form water. 
Water, consequently, is a Protoxide of Hydrogen. 

Upon the principle of atomic weights, — relative quantities, 
by weight, in which, the elements combine in forming compounds, 
based upon the standard already shown, — we have, for water, 
H 1 -j- O 8 = Aq. 9. That is, an atom of hydrogen is represented 
by 1, an atom of oxygen by 8, and an atom of water by 9. 

By the same rule as the preceding, the constituents of atmos- 
pheric air are found to be to each other, in volume, as 4 to 1 ; four 
volumes of nitrogen and one volume of oxygen make one volume 
of atmospheric air. The weight of nitrogen to hydrogen, per 
equal volumes, is as .972 to .0089, as 14.11 to 1. Atomically, 
therefore, it is as 7.055 to 1 ; hence, we have W -\- O = 36.22, 
the atomic weight of atmospheric air. 

The vast condensation of the gases which takes place, in some in- 
stances, in forming compounds, may be conceived of, and the process 
for determining the same exhibited by a single illustration. We will 
take, for example, water. A single cubic inch of distilled water, at 
60°, weighs 252.48 grains. Its weight is to that of dry atmosphere, 
at the same temperature, as 827.8 to 1. A cubic inch of dry atmos- 
phere, therefore, at that density, weighs .305 of a grain. Hydrogen, 
we find by the table of Specific Gravities, weighs .0689 as much 
as atmosphere, and oxygen 1.1025 as much. A cubic inch of hydro- 
gen, therefore, weighs .0689 X .305 = .0210145 of a grain, and 
a cubic inch of oxygen 1.1025 X -305, = .3362625 of a grain. 
The constituents of water by volume are 2 of the first mentioned gas 
to 1 of the latter; and .0210145 X 2 + -3362625 = .3782915 of a 
grain, = weight of three cubic inches of the uncondensed compound, 
} of which, .1260972 of a grain, is the weight of a volume 1 cubic 
inch. 

As the weight of a given volume of the uncondensed compound, is 
to the weight of an equal volume of the condensed compound, so are 
their respective volumes, inversely : then — 

.1260972 : 252.48 :: 1 : 2002.26, the number of cubic inches of the 
two gases condensed into 1 inch to form water ; a condensation of 
2001 times. Of this volume of gases, g, or 1334.84 cubic inches, is 
hydrogen ; the remaining third, 667.42 cubic inches, is oxveren. 

8* 



90 PROPERTIES, ETC., OF BODIES. 

The foregoing method, though strictly correct, does not exhibit in a 
general way the most expeditious for solving questions of that nature, 
the condensation which takes place in the gases on being converted 
into solids, or dense compounds. It was resorted to, in part, as a 
means through which to exhibit principles and proportions pertaining 
thereto. 

As before ; one cubic inch of water weighs 252.48 grains, ^ of 
which, or 28.05-f- grains, is hydrogen, and |. ? or 224. 43 — grains, is 
oxygen. The volume of 1 grain of oxygen is 2.97-}- cubic inches, and 
the volume of hydrogen is 1G times as much, or47.58-[- cubic inches. 
Therefore, 28.05 X 47.58 = 1334.62, and 224.43 X 2.97 = 665.56, 
safe 2001.18, condensation, as before. 



Properties of the simple substances, and some of their compounds, not 
given in the foregoing . 
Bromine, — at common temperatures, a deep reddish-brown vola- 
tile liquid ; taste caustic ; odor rank ; boils at 116° ; congeals at 4° ; 
exists in sea-water, in many salt and mineral springs, and in most 
marine plants ; action upon the animal system very energetic and 
poisonous — a single drop placed upon the beak of a bird destroys the 
bird almost instantly. A lighted taper, enveloped in its fumes, burns 
with a flame green at the base and red at the top ; powdered tin or 
antimony brought in contact is instantly inflamed ; potash is exploded 
with violence. 

Chlorine, — a greenish-yellow, dense gas ; taste astringent; odor 
pungent and disagreeable ; by a pressure of 60 lbs. to the square inch 
is reduced to a liquid, and thence, by a reduction of the temperature 
below 32°, into a solid. It exists largely in sea-water — is a constit- 
uent of common salt, and forms compounds with many minerals ; is 
deleterious, irritating to the lungs, and corrosive ; has eminent 
bleaching properties, and is the greatest disinfecting agent known ; 
a lighted taper immersed in it burns with a red flame ; pulverized 
antimony is inflamed on coming in contact, so is linen saturated with 
oil of turpentine ; phosphorus is ignited by it, and burns, while im- 
mersed, with a pale-green flame ; with hydrogen, mixed measure for 
measure, it is highly explosive and dangerous. 

Fluorine, — a gas, similar to chlorine, — exists abundantly in 
fluor-spar. 

Oxygen, — a transparent, colorless, tasteless, inodorous, innoxious 
gas ; supports respiration and combustion, but will not sustain life for 
any length of time, if breathed in a pure state. It is by far the most 
abundant substance in existence ; constitutes .* of the atmosphere ; 



PROPERTIES, ETC., OF BODIES. 91 

| of water ; and nearly the whole crust of the earth is oxidized sub- 
Btances. For further combinations and properties, see tables of Ele- 
mentary Constituents and Chemical Elements. ' « 

Iodine, — at common temperatures, a soft, pliable, opaque, bluish- 
black solid ; taste acrid ; odor pungent and unpleasant ; fuses at 225° ; 
boils at 347° ; its vapor is of a beautiful violet color ; it inflames 
phosphorus, and is an energetic poison ; exists mainly in sea-weeds 
and sponges. 

Hydrogen, — a transparent, colorless, tasteless, inodorous, innox- 
ious gas ; if pure, will not support respiration ; if mixed with oxy- 
gen, produces a profound sleep ; exists largely in water ; is the basis 
of most liquids, and is by far the lightest substance known ; burns in 
the atmosphere with a pale, bluish light ; mixed with common air, 1 
measure to 3, it is explosive ; mixed with oxygen, 2 measures to 1, 
it is violently so. 

Nitrogen, or Azote, — a transparent, colorless, tasteless, inodorous 
gas ; will not support respiration or combustion, if pure ; exists 
largely as a constituent of the atmosphere — in animals, and in fun- 
gous plants ; is evolved from some hot springs ; in connection with 
some bodies, appears combustible. 

Carbon, — the diamond is the only pure carbon in existence ; pure 
carbon cannot be formed by art ; charcoal is 97 per cent, carbon ; plum- 
bago, 95 ; anthracite, 93. Carbon is supposed by some to be the hard- 
est substance in nature. A piece of charcoal will scratch glass ; but 
it is doubtful if this is not due to the form of its crystals, rather than 
to the first mentioned quality. It is doubtless the most durable. For 
combinations, &c, see table. 

Boron, — a tasteless, inodorous, dark olive-colored solid. 

Silicon, — a tasteless, inodorous solid, of a dark-brown color; 
exists largely in soils, quartz, flint, rock-crystal, &c. ; burns readily 
in air — vividly in oxygen gas ; explodes with soda, potassa, barryta. 

Phosphorus, — a transparent, nearly colorless solid, of a wax- 
like texture ; fuses at 108°, and at 550° is converted into a vapor; 
exists mainly in bones — most abundant in those of man — is poison- 
ous ; at common temperatures it is luminous in the dark, and by fric- 
tion is instantly ignited, burning with an intense, hot, white flame ; 
must be kept immersed in water. 

Selenium, — a tasteless, inodorous, opaque, brittle, lead-colored 



92 PROPERTIES, ETC., OF BODIES. 

solid, in the mass ; in powder, a deep-red color ; becomes fluid at 
216°, boils at 050° ; vapor, a deep yellow ; exists but sparingly, 
mainly in. combination with volcanic matter ; is found in small quan- 
tities combined with the ores of lead, silver, copper, mercury. 

Ammoniacal gas, — N -f- H 3 ; transparent, colorless, highly pun- 
gent and stimulating ; alkaline ; is converted into a transparent liquid 
by a pressure of 6.5 atmospheres, at 50° ; does not support respira- 
tion ; is inflammable. 

Carbonic acid gas, — C -f- O 2 ; transparent, colorless, inodorous, 
dense ; is converted into a liquid by a pressure of 36 atmospheres ; 
exists extensively in nature, in mines, deep wells, pits ; is evolved 
from the earth, from ordinary combustion, especially from the combus- 
tion of charcoal, and from many mineral springs ; is expired by man 
and. animals; forms 44 per cent, of the carbonate of lime called mar- 
ble ; the brisk, sparkling appearance of soda-water, and most mineral 
waters, is due to its presence. It is neither a combustible nor a sup- 
porter of combustion ; and, when mixed with the atmosphere to an 
extent in which a candle will not burn, is destructive of life. Being 
heavier than atmosphere, it maybe drawn up from wells in large open 
buckets ; or it may be expelled by exploding gunpowder near the bot- 
tom. Large quantities of water thrown in will absorb it. 

The above gas is expired by man to the extent of 1632 cubic inches 
per hour ; it is generated by the burning of a wax candle to the ex- 
tent of 800 cubic inches per hour; and, by the burning of "Cam- 
phene," (in the production of light equal to that afforded by 1 wax 
candle,) to the extent of 875 cubic inches per hour. Two burning 
candles, therefore, vitiate the air to about the same extent as 1 per- 
son. 

Carbonic oxide gas, — C -J- O ; transparent, colorless, insipid; 
odor offensive ; does not support combustion ; an animal confined in 
it soon dies ; is highly inflammable, burning with a pale blue flamje ; 
mixed with oxygen, 1 to 2, is explosive — with atmosphere, even in 
small quantity, is productive of giddiness and fainting. 

Carbureted hydrogen gas, — C -f- H 2 ; transparent, colorless, taste- 
less, nearly inodorous ; exists in marshes and stagnant pools — is 
there formed by the decomposition of vegetable matter ; extinguishes 
all burning bodies, but at the same time is itself highly combustible, 
burning with a bright but yellowish flame ; it is destructive to life, if 
respired. 

Cyanogen — Bicarburet of Nitrogen — a gas, — N -{- C 2 ; trans 
parent, colorless, highly pungent and irritating ; under a pressure of 



PROPERTIES, ETC., OF BODIES. 93 

3.6 atmospheres, becomes a limpid liquid ; burns with a beautiful 
purple flame. 

Hydrochloric acid gas — Muriatic acid gas, — H -f - CI. (chlorine) ; 
transparent, colorless, pungent, acrid, suffocating ; strong acid taste. 

Nitrous oxide gas — Protoxide of Nitrogen, " laughing gas ," — 
N -j- O ; transparent, colorless, inodorous ; • taste sweetish ; powerful 
stimulant, when breathed, exciting both to mental and muscular ac- 
tion ; can support respiration but from 3 to 4 minutes ; is often per- 
nicious in its effects. 

Nitric oxide gas — Binoxide of Nitrogen, — N -J- O- ; transparent, 
colorless ; wholly irrespirable ; lighted charcoal and phosphorus burn 
in it with increased brilliancy. 

Olefiant gas — Bicarbureted hydr^jen gas — " coal gas," — C 2 -{- 
H 2 ; transparent, colorless, tasteless, nearly inodorous, when pure ; 
does not support respiration or combustion ; a lighted taper immersed 
in it is immediately extinguished. It burns with a strong, clear, 
white light ; mixed with oxygen, in the proportion of 1 volume to 3, 
it is highly explosive and dangerous. 

Phosphureted hydrogen gas, — P -j- H 3 ; colorless; odor highly 
offensive ; taste bitter ; exists in the vicinity of swamps, marshes, 
and grave-yards ; is formed by the decomposition of bones, mainly ; 
is highly inflammable ; takes fire spontaneously on coming in contact 
with the atmosphere ; mixed with pure oxygen, it explodes. It is 
the veritable " Will o' the wisp." 

Sulphureted hydrogen gas — Hydro sulphuric acid gas, — S -f- H ; 
transparent, colorless ; taste exceedingly nauseous ; odor offensive 
and disgusting ; is furnished by the sulphurets of the metals in gen- 
eral — also by filthy sewers and putrescent eggs. It is very destruc- 
tive to life ; placed on the skin of animals, it proves fatal. It burns 
with a pale blue flame, and, mixed with pure oxygen, it is explosive. 

Hydrocyanic acid — Prussic acid, — N -J- C 2 -j- H ; a colorless, 
limpid, highly volatile liquid; odor strong, but agreeable — similar 
to that of peach-blossoms ; it boils at 79° and congeals at ; exists 
in laurel, the bitter almond, peach and peach kernel. It is a most 
virulent poison, — a drop placed upon a man's arm caused death in a 
few minutes. A cat, or dog, punctured in the tongue with a needle 
fresh dipped in it, is almost instantly deprived of life. 

Hydrofluoric acid, — F -\- H ; a colorless liquid, in well stopped 
lead or silver bottles, at any temperature between 32° and 59°. It is 



94 PROPERTIES, ETC., OF BODIES. 

obtained by the action of sulphuric acid on fluor-spar. It readily 
acts upon and is used for etching on glass. It is the most destructive 
to animal matter of any known substance. 

Nitrohydrochloric acid — " aqua regia," — (1 part nitric acid and 4 
parts muriatic acid, by measure ;) — a solvent for gold. The best sol- 
vent for gold is a solution of sal ammoniac in nitric acid. 

Nitrosulphuric acid, — (1 part nitric acid and 10 parts sulphuric 
acid , by measure) — a solvent for silver ; scarcely acts upon gold, 
iron, copper, or lead, unless diluted with water ; is used for separat- 
ing the silver from old plated ware, &c. The best solvent for silver, 
and one which will not act in the least upon gold, copper, iron, or 
lead, is a solution of 1 part of nitre in 10 parts of concentrated sul- 
phuric acid, by weight, heated to 160°. This mixture will dissolve 
about £ its weight of silver. The silver may be recovered by adding 
common salt to the solution, and the chloride decomposed by the car- 
bonate of soda. 

Selenic acid, — Se -|- O 3 ; obtained by fusing nitrate of potassa with 
selenium — a solvent for gold, iron, copper, and zinc. 

Silicic acid, — (Silica — silex ; base Silicon) — Si — {— O 3 ; exists 
largely in sand. Common glass is fused sand and protoxide of potas- 
sium (carbonate of potassa — potash) in the proportion of 1 part by 
weight of the former to 3 of the latter. 

Manganese, compounded with oxygen, in different proportions, im- 
parts the various colors and tints given to fancy glass ware, now so 
generally in vogue. 

Butylene, — C 4 -|-H 4 ; sp. gr. 1.9348; a gaseous hydro-carbon 
derived from the distillation of coal-tar ; illuminating power, com- 
pared with that of olefient gas, as 2 to 1. 

Propylene, — C 3 -f-H 3 ; sp. gr. 1.4511 ; a gaseous hydro-carbon 
derived from tbe distillation of coal-tar ; illuminating power, com- 
pared with that of olefient gas, as 1.5 to 1. 

Napthaline Vapor, — C 10 -j- H 4 , tbe vapor of solidified olefient 
gas. 

Turpentine Vapor, — C 10 -j- H 8 . 



SECTION III. 
PKACTICAL ARITHMETIC. 



VULGAR FRACTIONS. 

A fraction is one or more parts of a Unit. 

A vulgar fraction consists of two terms, one written above the 
other, with a line drawn between them. 

The term below the line is called the denominator, as showing the 
denomination of the fraction, or number of parts into which the unit 
is broken. 

The term above the line is called the numerator, as numbering the 
parts employed. These together constitute the fraction and its 
value. 

A vulgar fraction always denotes division, of which the denomina- 
tor is the divisor and the numerator the dividend. Its value as a unit 
is the quotient arising therefrom. 

A simple fraction is either a proper or improper fraction. 

A proper fraction is one whose numerator is less than its denomina- 
tor, as £, f , § £, &c. 

An improper fraction has its numerator equal to or greater than its 
denominator, as f , ^, \^, &c. 

A mixed fraction is a compound of a whole number and a fraction, 
as li, 5i£, 12ft, &c. 

A compound fraction is a fraction of a fraction, as £ of f ; | of 
! of if, &c. 

A complex fraction has a fraction for its numerator or denom- 
I 4 i 51 
inator, or both, as -, -, |, — , &c, and is read \ -f- 3 ; 4 -5- § ; 
3' ■§■ ■-£ 4 

REDUCTION OF VULGAR FRACTIONS. 
To reduce a fraction to its lowest terms. 

This consists in concentrating the expression without changing the 
value of the fraction or the relation of its parts. 

It supposes division, and, consequently, by a measure or measures 
common to both terms. 

It is said to be accomplished when no number greater than 1 will 
divide both terms without a remainder : — therefore. 



96 VULGAR FRACTIONS. 

Rule. — Divide both terms by any number that will divide „nem 
without a remainder, and the quotient again as before ; continue so to 
do until no number greater than 1 will divide them, — or divide by 
the greatest common measure at once. 

Example. — Reduce T 8 5 6 -j%- to its lowest terms. 
4 ) #ft-*H *«*«# 4^9 = £}-5-3 = f Ans. 
To reduce an improper fraction to a mixed or whole number. 

Rule. — Divide the numerator by the denominator and to the 
whole number in the quotient annex the remainder, if any, in form of 
a fraction, making the divisor the denominator as before ; then reduce 
the fraction to its lowest terms. 

Example. $- 1* ; if=l-& = l*; ff = 2 - 
To reduce a mixed fraction to an equivalent improper fraction. 

Rule. — Multiply the whole number by the denominator of the 
fractional part, and to the product add the numerator, and place their 
sum over the said denominator. 

Example. — Reduce 3% and 12J- to improper fractions. 

3X4 = 12 + 1 = -^. Ans. 12 X 9 + 8 = H~- Ans - 

To reduce a whole number to an equivalent fraction having a given 
denominator. 

Rule. — Multiply the whole number by the given denominator, 
and place the said denominator under the product. 

Example. — How may 8 be converted into a fraction whose de- 
nominator is 12 1 

8 X 12 = f f . Ans. 

To reduce a compound fraction to a simple one. 
Rule. — Multiply all the numerators together for a numerator, and 
all the denominators together for a denominator ; the fraction thus 
formed will be an equivalent, but often not in its lowest terms. Or, 
concentrate the expression, when practicable, by reciprocally expung- 
ing, or writing out, such factors as exist or are attainable common to 
both terms, and then multiply the remaining terms as directed above. 

Note. — This last practice is called cancellation, or cancelling the terms. It consists, 
as has been stated, in reciprocally annulling, or casting out, equal values from both terms, 
whereby the expression is concentrated, and the relation of the parts kept undisturbed ; 
and it may always be carried to the extent of reducing the fraction to its lowest terms, 
before any multiplication, as final, is resorted to ; and often, therefore, to the extent that 
such multiplication is inadmissible, the terms having been cancelled by each other until 
but a single number is left in each. 



VULGAR FRACTIONS. 97 

Example. — Reduce § of | of £ to a simple fraction. 

Operation by multiplication, § X f X £ = ^ = £. Ans. 

$3 1 
Operation by cancellation, ^ . = £. Ans. 

Example. — Reduce § of f- of -^- of § of •§■ of 2 to a simple 
fraction. 

By multiplication, § X j- X - 1 /- X f X f X f = If f | = f. An*. 
The last example stated ) 2 3 12 6 5 2 
for cancellation, i 3 4 8 8 9 

PROCESS OF CANCELLING THE ABOVE. 

1. The 3 in num. equals the 3 in denom., therefore erase both. 

2. The first 2 in num. equals or measures the 4 in denom. twice, therefore place a 2 
under the 4, and erase the 4 and 2 which measured it — (as 4 : 2 : : 2 : 1.) 

3. The 2 (remaining factor of 4 and 2 erased) in denom., and the remaining 2 in num., 
will cancel each other, — erase them. 

4. The 12 and 6 in num. = 72, and the 9 and 8 in denom. = 72 ; these; therefore, in 
their relations as factors equal each other, and may be erased. 

The remaining factors represent the true value of the compound fraction, and will be 
found = |, as by multiplication. 

Example. — Reduce ^-f of ^ to a simple fraction. 

3 

3 

*ix_4. or/-^44(= ^l8 - 6 ' andl2 - 6 )=txT 7 3 



13 xU' '13 X a Ans . 



To reduce fractions of different denominations to an equivalent simple 
one, — to a fraction having a common denominator. 
Rule. — Multiply each numerator by all the denominators except 
its own and add the products together for the numerator, and multiply 
all the denominators together for a denominator. 

Note. — Whole numbers and fractions other than simple, must first be reduced to sim- 
ple fractions before they can be reduced to a fraction having a common denominator. 

Example. — Reduce § and f to an equivalent simple fraction. 

*X*-«- + •-■«: Ans. 
Example. — Reduce ^-, f, ■£, and ^ to an equivalent. 

9 



98 VULGAR FRACTIONS. 

To i educe a complex fraction to a simple one. 
Rule — Multiply the numerator of the upper fraction by the 
denominator of the lower, for the new numerator ; and the denomi- 
nator of ihe upper by the numerator of the lower for the new denom- 
inator. 

I 4 1 51 

Examples. — Reduce — , — , -f > and — - each to a simple fraction. 
3 f I 4 > 

2 ~ T — ¥ ' T ~ "5 U' 2 ~ 4 — 2 A 1U — ^> — 3 ' °2 — 

■fcL, and V- X J = -VS = If- Ans. 
To reduce Vulgar Fractions to equivalent Decimals. 

Rule. — Divide the numerator by the denominator ; the quotient is 
the decimal, or the whole number and decimal, as the case may be. 

Example. — Reduce |-, 4-| , ^#-, to decimals. 
7 -r- 8 = 0.875 ; 4f = -\ 3 -, == 4~ 6 ; 14 -i- 12 = 1.166 -f. Ans. 

To find the greatest common measure of two or more given numbers. 

Rule. — Divide the given numbers by any measure common 
to them all, and set the quotients in a line beneath ; then divide 
the quotients by any measure common to them, and set the quo- 
tients beneath ; and so on until the quotients are no longer com- 
mon multiples of any one number greater than unity; the product 
of all the divisors or common measures employed will be the 
greatest common measure. 

Example. — What is the greatest common measure of 84 and 
36 ? Also of 32, 24, and 16 ? Also of 182, 104, and 52 ? 



4)84.36 




8)32. 


24.16 






2)3 


L82.104.52 


3)21 .9 




4 


. 3.. 2.- 


-8. 


Ans. 


13) 


91. 


,52.26 


7.3- 


-4X3: 


= 12. 


Ans. 








7 


.4.2 



2.=26. Ans. 

Note. — When any number in the series is prime to either of the others, the 
numbers are collectively incommensurable ; that is to say, their greatest com- 
mon measure is unity, or 1. 

To find the least comm on multiple of two or more given numbers. 

Rule. — Divide all the given numbers that are commensurable 
with each other by any measure that is common to them, and set 
the quotients, together with the undivided numbers, if any, in a 
line beneath ; then divide the quantities in the second line as be- 
fore, and so on until no two quantities in the last line are common 
multiples of any number greater than unity, or 1 ; the product of 
all the common measures employed into the product of all the 
numbers in the last line will be the least common multiple of 
the given numbers. 



VULGAR FRACTIONS. 99 

Example. — What is the least common multiple of 27 and 36 ? 
Also of 182, 104, and 52 ? Also of 24, 14, 12, and 7 ? 
9) 27. 36 

3 . 4=4X3X9 = 108. 





2) 182 


104 


52 


Ans. 


13) 91 


. 52 


26 




2) 7 


. 4 


. 2 



7) 24. 


14. 


12 


7 


2) 24 


. 2 


12 


1 


6) 12 


. 1 


. 6 





2 . l.=728. Ans. 



2.1.1. 168. Ans. 

ADDITION OF VULGAR FRACTIONS. 

Sum of the products of each numerator with all the denominator except that of the 
numerator involved, forms numerator of sum. 
Product of all the denominators forms denominator of sum. 

Rule. — Arrange the several fractions to be added, one after 
another, in a line from left to right ; then multiply the numerator of 
the first by the denominator of the second, and the denominator of the 
first by the numerator of the second, and add the two products 
together for the numerator of the sum ; then multiply the two denom- 
inators together for its denominator ; bring down the next fraction, and 
proceed in like manner as before, continuing so to do until all the 
fractions have been brought down and added. Or, reduce all to a 
common denominator, then add the numerators together for the 
numerator of the sum, and write the common denominator beneath. 

Examples. — Add together \, §, f, and $?. 

I X I = I X I = f f Xi = W" = ¥- = 3J. Ans. 

= *£■ = 3 J. Ans. 

SUBTRACTION OF VULGAR FRACTIONS. 

Product of numerator of minuend and denominator of subtrahend, forms numerator of 
minuend, for common denominator. 

Product of numerator of subtrahend and denominator of minuend, forms numerator of 
subtrahend, for common denominator. 

Product of denominators forms common denominator. 

Difference of new found numerators forms the numerator, and common denominator the 
denominator, of the difference, or remainder sought. 

Rule. — Write the subtrahend to the right of the minuend, with 
the sign ( — ) between them ; then multiply the numerator of the 
minuend by the denominator of the subtrahend, and the denominator of 
the minuend by the numerator of the subtrahend ; subtract the latter 
product from the former, and to the remainder or difference affix tho 



100 VULGAR 'FRACTIONS. 

product of the two denominators for a denominator ; the sum thus 
formed is the answer, or true difference. 

Examples. — Subtract \ from f , also f from \^ . 

i|_| = 55- 5l = gV Ans . 

DIVISION OF VULGAR FRACTIONS. 

Product of numerators of dividend and denominators of divisor, forms numerator of 
quotient. 

Product of denominators of dividend and numerators of divisor, forms denominator of 
quotient; therefore, 

Rule. — Write the divisor to the right of the dividend with the 
sign (-7-) between them; then multiply the numerator of the dividend 
by the denominator of the divisor, for the numerator of the quotient, 
and the denominator of the dividend by the numerator of the divisor, 
for the denominator of the quotient. Or, invert the divisor, and mul- 
tiply as in multiplication of fractions. Or, proceed by cancellation, 
when practicable. 

Examples. — Divide J by j ; J by J ; | by |-J ; and \ of f of 
I of # by J of lof | off. 

i-H = t; |-4-i=i; *-*-«-»; or|><if = ||. Ans, 
ix|x$x*=Ak = A, and ixixix§=& = T V> 

and * + T \ = n = ¥ = 6|. Ans. 

FORM FOR CANCELLATION. EXAMPLE LAST GIVEN. 

1354 4243 20 

= -—. Ans., as above. 



2463 1132 3 



Note. — The foregoing example can be cancelled to the extent of leaving but a 4 and a 
5 (= 20) numerators, and a 3 denominator. Units, or l's, in the expressions, are value- 
less, as a sum multiplied by 1 is not increased. 

MULTIPLICATION OF VULGAR FRACTIONS. 

Product of numerators of multiplier and multiplicand, forms numerator of product. 
Product of denominators of multiplier and multiplicand, forms denominator of product. 

Rule. — Multiply the numerators together for a numerator, and the 
denominators together for the denominator. 

Examples. — Multiply J by \ ; § by 7 ; \l by \ 2 - ; J of § of } 
by j of \ off. 

ix j = j; fx T = ¥; ^x-v 2 -=-W-=-V-; ix §x| 
= A = i, and | x | X § = A = i> and i X i = T V Ans - 



VULGAR FRACTIONS. 101 

MULTIPLICATION AND DIVISION OF FRACTIONS, COMBINED. 

It has been seen that a compound fraction is converted into an 
equivalent simple one, by multiplying the numerators together for a 
numerator, and the denominators together for a denominator; and 
it has also been seen that a series of simple fractions are con- 
verted into a product, by the same process. It is therefore evident 
that compound fractions and simple, or a series of compound and a 
series of simple, may be multiplied into each other, for a product, by 
multiplying all the numerators of both together for a numerator, and 
all the denominators of both together for a denominator ; and that the 
product will be the same as would be obtained, if the compound were 
first converted into an equivalent simple fraction, and the simple frac- 
tions into a product or factor, and these multiplied together for a 
product. 

It has also been seen that a fraction is divided by a fraction by mul- 
tiplying the numerator of the dividend by the denominator of the 
divisor, for the numerator of the quotient, and the denominator of the 
dividend by the numerator of the divisor, for the denominator of the 
quotient ; and that this multiplication becomes direct as in multiply- 
ing for a product, if the divisor is inverted. And it is clear that a 
compound divisor, or a series of simple divisors, or both, may be used 
instead of their simple equivalent, and with the same result, if all are 
inverted. 

It is therefore evident that any proposition, or problem, in fractions, 
consisting of multiplications and divisions both, and these only, no 
matter how extensive and numerous, or whether in compound frac- 
tions, or simple, or both, may be solved, and the true result obtained, 
as a product, by simply multiplying all the numerators in the state- 
ment together for a numerator, and all the denominators in the state- 
ment for a denominator, all the divisors in the statement being 
inverted ; that is, all the numerators of the divisors being made denom- 
inators in the statement, and all the denominators of the divisor being 
made numerators in the statement. And it is further evident that a 
proposition stated in this way, admits of easy cancellation as far as 
cancellation is practicable, which is often to great extent. 

Example. — It is required to divide 12 by f of | ; to multiply the 
quotient by the product of 4 and 8 ; to divide that product by |- of | 
of 8 ; to multiply the quotient by £ of §• of T 9 ^- ; and to divide tha* 
product by the product of 5 and 9. 

9* 



102 



VULGAR FRACTIONS. 



STATEMENT. 
(Dividends read from right to left, divisors from left to right.) 



Numerators of dividends and denominators of divisors. 




$ Dividend of 
? statement. 



Denominator 
of statement. 



i>coQOcoco^io05 c Divisor of 
/ statement. 



•sjosiAipjo sjoiEjgumu pire spuapiAip JO SJOlBUllUOtOQ 



The answer to the above proposition is 1J-J-, and the proposition 
as stated may be readily cancelled to its lowest terms. It may be 
cancelled to the extent of leaving but 4, 4, 2 in the numerator, and 7, 
3, in the denominator, ^J^- 2 - = f f = 1 J-|. 

To reduce a fraction in a higher denomination to an equivalent fraction 
in a given lower denomination. 
Rule. — Multiply the fraction to be reduced — numerators into 
numerator and denominators into denominator — by a fraction whose 
numerator represents the number of parts of the lower denomination, 
required to make one of the denomination to be reduced. 

Example. — Reduce £ of a foot to an equivalent fraction in inches. 
iX \ 2 - = -¥- = -¥• An*. 

Example. — Reduce |- of a pound to an equivalent fraction in § 
ounces. 

| X -V 6 - = -V- 4- | = W = '¥• Ans. 
Or, | X -V 6 - X I = W = 2 T°- A*"- 

To reduce a fraction in a lower denomination to an equivalent fraction 
in a given higher denomination. 
Rule. — Multiply the fraction to be reduced — numerator into 
denominator and denominator into numerator — by a fraction whose 
numerator represents the number of parts required of the lower 
denomination to make 1 of the higher. 

Example. — Reduce %£■ inches to an equivalent fraction in feet. 
V + If. = || =- J. Arts. Or, V" X T V - f i - h ***> 



VULGAR FEACTIONS. 103 

Example. — Reduce &£- two third ounces to an equivalent frac- 
tion in pounds. 

40 y 2 80 J_ 16 — 80 = 5 Arts 

Or,-V-XlXT 1 F = fl = t- ***• 

To reduce a fraction in a higher to whole numbers in lower denomi 
nations. 
Rule. — Multiply the numerator of the given fraction by the num- 
ber of parts of the next lower denomination that make one of the 
given fraction, and divide the product by the denominator. Multiply 
the numerator of the fractional part of the quotient thus obtained by 
the number of parts in the next lower denomination that make 1 of 
the denomination of the quotient, and divide by its denominator for 
whole numbers as before ; so proceed until the whole numbers in each 
denomination desired are obtained. 

Example. — How many hours, minutes, and seconds, in -^ of a 
day? 

^X24^1_6 =1 5 ) 3X60=l|0 =s25j 5X60=3flO = 42 6 ) = 

15 h., 25 m., 42f- sec. Ans. 
Example. — How many minutes in T 9 ^ of a day? 

_9 X 2 4 X 6 O =i2_9_6J> = 925 5 . Ans% 

To reduce fractions, or whole numbers and fractions , in lower denomi- 
nations, to their value in a higher denomination. 
Rule. — Reduce the mixed numbers to improper fractions, find 
their common denominator, and change each whole number and 
numerator to correspond therewith. Then reduce the higher numbers 
to their values in the lowest denomination, add the value in the lowest 
denomination thereto, and take their sum for a numerator. Multiply 
the common denominator by the number required of the lowest denom- 
ination to make one of the next higher, that product by the number 
required of that denomination to make 1 of the next higher, and so 
on, until the highest denomination desired is reached, and take the 
product for a denominator, and reduce to lowest terms. 

Example. — Reduce 5| oz., 3-i dwts., 2^ grs., troy, to lbs. 
Ig-.lg-. | = ±&°£6 J L5 . therefore, 

160 X 20 = 3200 
96 



3296 X 24 = 79104 

75 



79179 
30 X 24 X 20 X 12 = 172600 



= .458 + lbs. Am. 



104 DECIMAL FRACTIONS. 

Example. — Reduce 11 hours, 59 minutes, 60 seconds, to the frac- 
tion of a day. 

11 X 60 = 660 
59 

719 X 60 = 43140 
60 



43200 ) 

> — £. Ans. 

= 86400 ) 



60 X 60 X 24 = 86400 

Example. — Reduce 15 h., 25 m., 42f- sec, to the fraction of a 
day. 

15 X 60 X 60 = 54000 
25 X 60=: ]500 

42f 



55542f 

7 

388800 



7 X 60 X 60 X 24 = 604800 

To work fractions, or whole numbers and fractions, by the Rule of 
Three, or Proportion. 

Rule. — Reduce the mixed terms to simple fractions, state the 
question as in whole numbers, invert the divisor, and multiply and 
divide as in whole numbers. 

Example. — If 2£ yards of cassimere cost $4|, what will | of a 
"ard cost? 2 J = | ; A\ = - 1 /- ; then, 

k • JL7_ • • 3. • = JL7.X 3 x 2 _ J 02 __ «(bi 07 k /[„<. 



DECIMAL FRACTIONS. 

A decimal fraction is written with its numerator only. Its denomi- 
nator is understood. It occupies one or more places of figures, and 
has a point or dot (.) prefixed or placed before it. The dot (.) alone 
distinguishes it from an integer or whole number. It supposes a 
denominator whose value is a unit broken into parts, having a ten- 
fold relation to the number of places the numerator occupies. The 
denominator, therefore, of any decimal is always a unit (1) with as 
many ciphers annexed as the numerator has places of figures. Thus, 
the denominator of .1, .2, .3, &c, is 10, and the fractions are read, 
one tenth, two tenths, three tenths, &c. The denominator of .01, .11, 
.12, &c, is 100, and these are read, one hundredth, eleven hundredths, 



DECIMAL FRACTIONS. 105 

twelve hundredths, &c. The denominator of .001, .101, .125, &c, is 
1000, and these are read one thousandth, one hundred and one thousandths, 
one hundred and twenty-five thousandths, &c. The denominator of a 
decimal occupying four places of figures as .7525 is 10000, and so on 
continually. 

The first figure on the right of the decimal point is in the place of 
tenths, the second in the place of tenths of tenths, or hundredths, the 
third in the place of tenths of tenths of tenths, or thousandths, &e. 
Thus the value of a decimal occupying four places of figures, as 

7525 ,7524 15k U~~~i 

•7525, for example, is , = , = , = —=- A — = 

___!__ 10000 1000 100 10 ~ 100 

3 J. 

1 — . A decimal is converted into a vulgar fraction of 

1 ' 100 * 

equal value, by affixing its denominator. 

Ciphers placed on the right of decimals do not change their value. 
Thus, .1850 = .185, plainly for the reason that the denominator of 
the latter bears the same relation to that of the former that 185 bears 
to 1850 ; from both terms of the fraction a ten fold has been dropped. 

Ciphers placed on the left of decimals decrease their value ten fold 
for every cipher so placed. Thus, .1 = T \j, .01 = T tnr> »001 = 
Tinny > & c - 

A mixed number is a whole number and a decimal. Thus, 4.25 is 
a mixed number. Its value is 4 units, or ones, and ■££$ of 1, = 
4j-^ = 4£. The number on the left of the separatrix is always a 
whole number — that on its right, always a decimal. 

ADDITION OF DECIMALS. 
Rule. — Set the numbers directly under each other according to 
their values, whole numbers under whole numbers, and decimals un- 
der decimals ; add as in whole numbers, and point off as many places 
for decimals in the sum as there are figures in that decimal occupying 
the greatest number of places. 

Examples. — Add together .125, .34, .1, .8672. Also, 125, 34.* 
.235. 1.4322. 



.125 125. 



.34 
.1 

.8672 
1.4322 Ans. 



34.11 
.235 
1.4322 
16077772 Ans. 



SUBTRACTION OF DECIMALS. 



Rule. — Set the numbers, the less under the greater, and in other 
resuects as directed for addition ; subtract as in whole numbers, and 



106 



DECIMAL FRACTIONS. 



point off as many places for decimals in the remainder as the decimal 
having the greatest number of figures occupies places. 

Examples. — Subtract .2653 from .8. Also, 11.5 from 238.134. 



.8 

.2653 

.5347 Ans. 



238.134 

11.5 
226.634 Ans. 



MULTIPLICATION OF DECIMALS. 

Rule. — Multiply as in whole numbers, and point off as many 
places for decimals in the product as there are decimal places in the 
multiplicand and multiplier both. If the product has not so many 
places, prefix ciphers to supply the deficiency. 

Examples. — Multiply 14.125 by 3.4. Also, 5.14 by .007. 



14.125 
3.4 
56500 
42375 
48.0250 = 48.025. Aits. 



5.14 
.007 



.03598 Ans. 



Note. — Multiplying by a decimal is equivalent to dividing by a whole number that 
bears the same relation to a unit that a unit bears to a decimal. Multiplying by a deci- 
mal, therefore, is equivalent to dividing by the denominator of a fraction of equal value 
whose numerator is 1, or of dividing by the denominator of a fraction of equal value whose 
numerator is more than I, and multiplying the quotient by the numerator. Thus, the 
decimal .25 = T 2 ^- = i, and the decimal .875 = y^V = i- And 14.23 X -25 =* 
3.5575, and 14.23 -r 4 = 3.5575. So, also, 14.23 X -875 = 12.45125, and 14.23 -j- 8 = 
1.77875 X 7 = 12.45125. It is sometimes a saving of labor and matter of convenience to 
achieve multiplication by this process. 

DIVISION OF DECIMALS. 

Rule. — Write the numbers as for division of whole numbers, then 
remove the separatrix in the dividend as many places of figures to the 
right, (supplying the places with ciphers if they are not occupied,) as 
there are decimal figures in the divisor ; consider the divisor a whole 
number and divide as in division of whole numbers. 



Example. — Divide .5 by .17. Also, .129 by 4. 



.17). 50(2.94+. Ans. 
34 
160 
153 

70 

68 

2 



4).129(.032-j-. 
12 
9 
8_ 
1 



Ans. 



DECIMAL FRACTIONS. 



107 



Examples.— Divide 16.5 by 1.232. Also, 1.2145 by 12.231. 



1.232,) 16.500, (13.3928-f-. Ans. 
1232 
4180 
3696 
4840 
3696 



11440 

11088 



12.231,)1.214,50(.09929-H Ans 
1 100 79 .0993— 5 JLns ' 



113 710 
110 079 
3 6310 
2 4462 
1 18480 
1 10079 
8401 



3520 
2464 
10560 
9856 

704 

Note. — Dividing by a decimal is equivalent to multiplying by a whole number that 
bears the same proportion to a unit that a unit bears to the decimal. Dividing by a deci- 
mal, therefore, is equivalent to multiplying by the denominator of a fraction of equal 
value whose numerator is 1, or multiplying by the denominator of a fraction of equal 
value whose numerator is more than 1, and dividing the product by the numerator. Di- 
viding by a fraction is equivalent to multiplying by it3 denominator and dividing the 
product by its numerator, or dividing by its numerator and multiplying the quotient by 

and .75 = J^„ = 3. And 12.24 -^ .5 = 24.48, 



its denominator. Thus, .5=-=-^ = 



7 5 a 

To-rj — *• 
and 12.24 X 2 = 24 - 48 - So > als0 - 12 - 24 -r -^ = 16. 32, and 12.24 X 4 = 4S - 96 t3 = 
16.32. This method of accomplishing division may often be resorted to with convenience. 



REDUCTION OF DECIMALS. 

To reduce a decimal in a higher to whole numbers in successive lower 
denominations. 

Rule. — Multiply the decimal by that number in the next lower 
denomination that equals one of the denomination of the decimal, and 
point off as many places for a remainder as the decimal so multiplied 
has places. Multiply the remainder by the number in the next lower 
denomination that equals 1 of the denomination of the remainder, and 
point off as before ; so continue, until the reduction is carried to the 
lowest denomination required. 

Example. — What is the value of .62525 of a dollar? 
.62525 



100 



Cents, 
Mills, 



62.52500 
10 



5.25000 An.. 62 cents 5£ mills. 



108 



DECIMAL FRACTIONS. 



Example. — What is the value of .46325 of a barrel? 

.46325 
32 



Gallons, 


14.82400 
4 


Quarts, 


3.296 
2 


Pints, 


.592 
4 



Gills, 2.368. Ans. 14 gals. 3 qts. 2 T 3 ^ gills. 
Example. — How many pence in .875 of a pound 1 ? 
.875X240=210. Ans. 

To reduce decimals, or whole numbers and decimals, in lower denomin- 
ations, to their value in a higher denomination. 
Rule. — Reduce all the given denominations to their value in the 
lowest denomination, then divide their sum by the number required of 
the lowest denomination to make one of the denomination to which 
the whole is to be reduced. 

Example. — Reduce 14 gallons, 3 quarts, 2.368 gills, to the deci- 
mal of a barrel. 

14 X 4 = 56 + 3 = 59 X 8 = 472 + 2.368 = 474.368. 
8 X 4 X 32 = 1024 ) 474.368 ( .46325. Ans. 

To work decimals, or whole numbers and decimals, by the Rule of 
Three, or Proportion. 
Rule. — State the question and work it as in whole numbers, 
taking care to point off as many places for decimals in the product to 
be used as the dividend, as there are decimals in the two terms which 
form it, and to remove the decimal point therein as many places to the 
right as there are decimals in the term to be used as a divisor, before 
the division is had. 

Example. — If .75 of a pound of copper is worth .31 of a dollar 
how much is 3.75 lbs. worth? 

.75 : .31 :: 3.75 
.31 
375 
1125 



.75) 1.16,25 ($1.55. Ans. 



PROPORTION. 109 

» 

PROPORTION, OR RULE OF THREE. 

The Rule of Proportion involves the employment of three terms 
— a divisor and two factors for forming- a dividend — and seeks a 
quotient, which, when the proposition is written in ratio, bears the 
same relation to the third term that the second term bears to the first. 
Two of the terms given are of like name or nature, and the other is 
of the name or nature of the quotient or answer sought. That of 
)he nature of the answer is always one of the factors for forming the 
dividend, and, if the answer is to be greater than that term, the larger 
of the remaining two is the other ; but if the answer is to be less 
than that term, the less of the remaining two is the other — the 
remaining term is the divisor. 

Example. — If $12 buy 4 yards of cloth, how many yards will 
$108 buy? 

£ X 108 _ 108 _ 

-jrx — 36 yards. Ans. 

3 

Example. — If 4 yards of cloth cost $12, how many dollars will 
36 yards cost? 

^--= 108 dollars. Ans. 
4 

Example. — If 30 men can finish a piece of work in 12 days, how 
many men will he required to finish it in 8 days 1 

30 X 12 A* A 

== 45 men. Ans. 

o 

Example. — If 45 men require 8 days to finish a piece of work, 
how many men will finish the same work in 12 days ? 
45 X 8 



12 



= 30 men. Ans. 



Example. — If 8 days are required by 45 men to finish a piece 
of work, how many days will be required by 30 men to finish the 
same work? 

8 X 45 

— 2q — = 12 da Y s - Ans. 

Example. — If 12 days are required by 30 men to perform a piece 

of work, how many days will be required by 45 men to do the same 

work? 

12X30 

— — = 8 days. Ans. 

45 J 

Example. — I borrowed of my friend $150, which I kept 3 months, 

and, on returning it, lent him $200 ; how long may he keep the sura 

10 



110 



COMPOUND PROPORTION. 



that the interest, at the same rate per cent. , may amount to that which 
his own would have drawn ? 

150 X 3 ~- 200 = 2-1 months. Ans. 

Example. — A garrison of 250 men is provided with provisions for 
30 days, how many men must be sent out that the provisions may last 
those remaining 42 days 1 

250 X 30 -i- 42 = 179, and 250 — 179 = 71. Ans. 

Example. — If to the short arm of a lever 2 inches from the ful- 
crum there be suspended a weight of 100 lbs., what power on the 
long arm of the lever 20 inches from the fulcrum will be required to 
raise it? 

20 : 2 :: 100= 10 lbs. Ans. 

Example. — At what distance from the fulcrum on the long arm of 
a lever must I place a pound weight, to equipoise or weigh 20 lbs., 
suspended 2 inches from the fulcrum at the other end 1 
1 : 2 :: 20 : 40 inches. Ans. 

Note. — If we examine the foregoing with reference to the fact, we shall see that every 
proposition in simple proportion consists of a term and a half! or, in other words, of a 
compound term consisting of two factors, and a factor for which another factor is sought 
that together shall equal the compound. We have only to multiply the factors of the 
compound together — and a little observation will enable us to distinguish it — and divide 
by the remaining factor, and the work is accomplished. See Compound Proportion. 



COMPOUND PROPORTION, OR DOUBLE RULE OF THREE. 

Compound Proportion, like single proportion, consists of three 
terms given by which to find a fourth — a divisor and two factors for 
forming a dividend — but unlike single proportion, one or more of the 
terms is a compound, or consists of two or more factors; and some- 
times a portion of the fourth term is given, which, however, is always 
a part of the divisor. 

Of the given terms, two are suppositive, dissimilar in their natures, 
and relate to each other, and to each other only ; and upon their rela- 
tion the whole is made to depend ; the remaining term is of the nature 
of one of the former, and relates to the fourth term, which is of the 
aature of the other. 

The object sought is a number, which, multiplied into the factor or 
factors of the fourth term given, if any, and if not, which of itself, 
bears the same proportion to the dissimilar term to which it relates, 
as the suppositive term of like nature bears to the term to which it 
relates. 

Rule.« — Observe the denomination in which the demand is made, 
and of the suppositive terms make that of like nature the second, and 
the other the first ; ma 1 e the remaining term the third term ; and, if 



COMPOUND PROPORTION. Ill 

there are any factors pertaining" to the fourth terra, affix them to the 
first ; multiply the second and third terms together and divide by the 
first, and the quotient is the answer, term, or portion of a term, 
sought. 

Example. — If 12 horses in 6 days consume 36 bushels of oats, 
how many bushels will suffice 21 horses 7 days 2 
12 x 6 : 36 :: 21 x 7 : ^. 

30 X 21 X 7 147 

=~i -x — = — — = 73£ bushels. Ans. 

1% X 2 

2 

Example. — If 12 horses in 6 days consume 36 bushel* of oats, 
how many horses will consume 73£ bushels in 7 days ? 

36 : 12 x 6 :: 73^ : 7 x *• 
12 X 6 X 73£ 147 
36 X 7 = T" = 21 h0rSeS * AnS ' 

Example. — If the interest on $1 is 1.4 cts. for 73 days, (exact 
interest at 7 per cent.,) what will be the interest on $150.42 for J 46 
days? 

73 : 1.4 :: 150.42 X 146 lx. 

1.4 X 150.42 X 146 

— : -- = $4.21. Ans. 

Example. — If the interest on $1 is 1.2 cts. for 73 days, (exact 
interest at 6 per cent.,) what will be the interest on $125 for 90 
days? 

73 : 1.2 :: 125 X 90 : x = $1.85. Ans. 

Example. — If $100 at 7 per cent, gain $1.75 in 3 months, hovr 
much at 6 per cent, will $170 gain in 11£ months? 

100 x 7 X 3 : 1.75 :: 170 x 6 x H-5 : x. 

1.75 X 170 X 6 X 11-5 -r- 100 X 7 X 3 = $9.77,5. Ans. 

Example. — By working 10 hours a day 6 men laid 22 rods of wall 
in 3 days ; how many men at that rate, who work but 9 hours a day, 
will lay 40 rods of wall in 8 days ? 

22 : 6 x 3 X 10 :: 40 : 9 x 8 X *. 
6 X 3 X 10 X 40 -r 22 X 9 X 8 = 4 T 6 T . Ans. 
Example. — If it costs $112 to keep 16 horses 30 days, and it 
costs as much to keep 2 horses as it costs to keep 5 oxen, how much 
will it cost to keep 28 oxen 36 days 1 # 



112 CONJOINED PROPORTION, OR CHAIN RULE. 

16 x 30 : 112 ::f X28X26:«. 

Or, — 16 X 30 X 5 : 112 :: 28 X 36 x 2 : x. 
7 1-2 

m 28 g|j = wxa x_t = 08 4m 

10 30 5 5X5 

*£ 

5 

Example. — If 24 men, in 8 days of 10 hours each, can dig a 
trench 250 feet long, 8 feet wide, and 4 feet deep, how many men, in 
12 days of eight hours each, will be required to dig a trench 80 feet 
long, 6 feet wide, and 4 feet deep? 
250 X 8 X 4 : 24 X 8 X 10 :: 80 X 6 X 4 : 12 X 8 X *= 5—. Am. 

Example. — If 120 men in six months perform a given task, work- 
ing 10 hours a day, how many men will be required to accomplish a 
like task in 5 months, working 9 hours a day ? 

120 X 6 X 10 = 5 X 9 X x. 
Or, — 1 : 120 X 6 X 10 » 1 : 5 X 9 X *• = 160. Ans. 

Example. — The weight of a bar of wrought iron, 1 foot in length, 
1 inch in breadth, and 1 inch thick, being 3.38 lbs., (and it is so,) 
what will be the weight of that bar whose length is 12£ feet, breadth 
3| inches, and thickness f of an inch ? 

1 : 3.38 :: 12.5 X 3.25 X .75 : x. 
Or, — 1 : 3.38 :: -^ X -^ X f : x, and 
3.38X25 X 13X3 = 10 bs 

2X4X4 ^ 

Example. — The weight of a bar of wrought iron, one foot in length 
and 1 inch square, being 3.38 lbs., what length shall I cut from a bar 
whose breadth is*2| inches, and thickness £ inch, in order to obtain 

10 lbs. ? 3.38 : l :: 10 : jyi x J X *• 

1 X 10 X 4 X 2 
3.38 X 11 X 1 = * ^ ** inCheS ' AnS - 



CONJOINED PROPORTION, OR CHAIN RULE. 

The Chain Rule is a process for determining the value of a given 
quantity in one denomination of value, in some other given denomi- 
nation of value ; or the immediate relationship which exists between 
two denominations of value, by* means of a chain of approximate steps, 



CONJOINED PROPORTION, OR CHAIN RULE. 113 

circumstances, or equivalent values, known to exist, which connect 
them. In every instance at least five terms or values are employed 
in the process, and in all instances the number employed will be un- 
even. A proposition involving but three terms, of this nature, is a 
question in single proportion. The equivalent values employed are 
divided into antecedents and consequents, or causes and effects ; and the 
value or quantity for which an equivalent is sought, is called the odd 
term. 

Rule. — 1. When the value in the denomination of the first antece- 
dent is sought of a given quantity in the denomination of the last conse- 
quent. — Multiply all the antecedents and the odd term together for a 
dividend, and all the consequents together for a divisor; the quotient 
will be the answer or equivalent sought. 

Rule. — 2. When the value in the denomination of the last consequent 
is sought of a given quantity in the denomination of the first antecedent. 
— Multiply all the consequents and the odd term together for a divi- 
dend, and all the antecedents together for a divisor ; the quotient will 
be the answer required. 

Example. — I am required to give the value, in Federal money, oi 
5 Canada shillings, and know no immediate connection or relationship 
between the two currencies — that of Canada and that of the United 
States. The nearest that I do know is that 20 Canada shillings have 
a value equal to 32 New York shillings, and that 12 New York shil- 
lings equal in value 9 New England shillings, and that 15 New Eng- 
land shillings -equal $2.50 ; and with this knowledge will seek the 
value, in Federal money, of the 5 Canada shillings. 

2^_X_9_X32X5 = $1. An.. 
15 X 12 X 20 

Example. — If $2£ equal 15 New England shillings, and nine shil- 
lings in New England equal 12 shillings in New York, and 32 shil- 
lings h\ New York equal 20 shillings in Canada, how many shillings 
in Canada will equal $1 ? 

2 3 * 

Example. — If 14 bushels of wheat weigh as much as 15 bushels 
of fine salt, and 10 bushels of fine salt as much as 7 bushels of coarse, 
and 7 bushels of coarse salt as much as 4 bushels of sand, how many 
bushels of sand will weigh as much as 40 bushels of wheat 1 

15 X 7 X 4 X 40 ,-,.,', A 

— — — — — =s 17-f bushels. Ans. 

14 X 10 X 7 T 

10* 



114 PERCENTAGE. 



PERCENTAGE. 

Pure percentage, or percentage, is a rate by the hundred of a 
part of a quantity or number denominated the principal, or basis. 
But percentage, considered as a means, and as commonly applied, 
is mixed and related in an eminent degree ; and in this light may 
be regarded as divided into orders bearing different names. 

Thus Interest is percentage related to intervals of time in the 
past. 

Discount is percentage related to interest, and intervals of time 
in the future. 

Profit and Loss is comparative percentage, or percentage related 
to the positive and negative interests in business, etc., etc. 

Pure percentage is commonly called brokerage when paid to 
a broker for services in his line. 

It is called commission when paid to or received by a factor 
or commission merchant for buying or selling goods. 

It is called premium by an insurance company, when taken for 
insuring against loss. 

It is called primage when it is a charge in addition to the 
freight of a vessel, etc. 

Comparative percentage relates to the differences of quantities, 
and is confined always to the idea of more or less. It implies ratio. 
This description of percentage, though much in practice, seems not 
to be well understood ; and often a quantity is indirectly stated to 
be many times less than nothing, or many times greater than it is. 
The difference of two quantities cannot be as great as a hundred 
per cent, of the greater, however widely unequal the quantities 
may be, nor as small as no per cent, of the greater or lesser, how- 
ever nearly equal they may be. No quantity or number can be as 
small as 1 time less than another quantity or number ; and there- 
fore cannot be as small as 100 per cent. less. But, since one quan- 
tity may be many by 1 time, or many times greater than another 
with which it is compared, it may be said to be many by 100 times, 
or many hundred per cent, greater. 

When one of two quantities in comparison is stated to be three 
times less, or three hnndred per cent, less, for instance, than the 
other, the expression is incorrect and absurd. The meaning evi- 
dently is, that it is two-thirds less, or only one-third as large as the 
other, — that it is 66f per cent, less, or only 33^- per cent, as large 
as the other. In common comparison, 1 is the measuring unit. In 
percentage, 100 is the measuring unit. 



PERCENTAGE. 115 

Let a = principal. 
b zzz percentage. 

s — amount (sum of the principal and percentage). 
d = difference of the principal and percentage. 
r = rate of the percentage. 
p — rate per cent, of the percentage. 

a = s — b=zb -± r =z 100b -±-p = 100s + (100 +p), 

b = s — a=zarz=zap -7- 100, 

p = lOOr =1006 -f«= 100(s — o) -T* Oi 

r = j9 -7- 100 = & -J- a z= (s — a) -7- a, 

5 = a -f- & = a(l + r) = a(100 -f- j?) -7- 100, 

d = a — &=2a — s = s — 2& = a(l — r). 

To find the Percentage. 

EXAMPLES. 

What is I of 1 per cent, of $200 ? 

6 = ar = op -f- 100 = $0.50. ./ircs. 
8 of 2 per cent, of 50 is what part of 50 ? 

50X8X2 ■ 

- 7T7T- = 14- Ans. 

7 X 100 y 

What is I of I of 1 of 24 per cent, of 150 lbs. ? 

150 X 12-r-100 = 18 lbs. Ans. 

What is 2| per cent, of 1 9 bushels ? 

J_9 x tWr= 0.45125 bushels. Ans. 

Bought a job lot of merchandise for $850, and sold it the same 
day, brokerage, 1\ per cent., for $975 ; what was the net gain? 
s — sr — a = s — (sr-\-a)=s(l — r) — a= 975 — 975 X -025 

— 850 = $100,625, Ans. 

To find the Rate or Rate Per Cent. 

EXAMPLES. 

What per cent, of $20 is $2 ? 

r = & -7- a, ^> =: 100b -±- a = 10 per cent. Ans. 
12 dozen is equal to what per cent, of 2 dozen ? 
12-7-2 = 6, 600 per cent. Ans. 



116 PERCENTAGE. 

What part of 5£ lbs. is f of 2 lbs. ? 

r = | X t 2 t = if = 0.27 T \. Ans. 

24| per cent, is what per cent, of 36| per cent. ? 

66f per cent. Ans. 

For an article that cost $4, $5 were received ; what per cent 
of $4 was received ? 

p = 5 X 100 -J- 4 = 125 per cent. ^4ns. 

A farmer sowed 4 bushels of wheat, which produced 48 bushels ; 
what per cent, was the increase ? 48 is more than 4 by what per 
cent, of 4 ? The difference of 48 and 4 is what per cent, of 4 ? 

a — b a , 100(a — 6) 48 — 4 jo . A 

r = ^— = T — 1, p=z i 1— = 48-i-4 — l = 

b b * p 4 

100(48 — 4) -^-4 = 1100 percent. Ans. 

What per cent, would have been the decrease, if he had sowed 
48 bushels, and harvested only 4 bushels ? 4 is less than 48 by 
what rate of 48 ? The difference of 48 and 4 is what per cent, of 

48? 

r=.(a — &) -f- a = 1 = 0.91f , or 91f per cent. Ans. 

Since water is composed of 8 atoms of oxygen and 1 atom of 
hydrogen, what per cent, of it is oxygen ? 8 is what per cent, 
of the sum of 8 and 1 ? 

a „ b 100a 8 nnnn 

or 88.89 - per cent. Ans. 

What per cent, of it is hydrogen ? 1 is what per cent, of the 
sum of 8 and 1 ? 

a b 100b 1 

r = i r-i — — rr> .P — — r~r = o ■ -, =.1111+, or 

a-\-b a-\-b r a-\-b 8-J-l '' 

11.11 -f- per cent. Ans. 

How many volumes of water must be added to 100 volumes of 
90 per cent, alcohol to reduce it to 50 per cent, alcohol or common 
proof? 90 is more than 50 by what per cent, of 50 ? The differ- 
ence of 90 and 50 is what per cent, of 50 ? 

(a — 5)100 (90 — 50)100 OA . 
P=- V— = 50 = 8 °- AnS - 



i»= * i Ti = ^ S n_^n = 33 i' Ans ' 



PERCENTAGE. 117 

How many volumes of 50 per cent, alcohol must be added to 
100 volumes of 90 per cent, alcohol to produce 80 per cent, alcohol ? 
90 is more than 80 by what per cent, of the difference of 80 and 
50 ? The difference of 90 and 80 is what per cent, of the differ- 
ence of 80 and 50 ? 

(a _ 5)100 __ (9Q — 80)1 00 
o — V ~~ 80 — 50 

How many volumes of 90 per cent, alcohol must be added to 1 00 
volumes of 50 per cent, alcohol to raise it to 80 per cent, alcohol ? 
50 is less than 80 by what per cent, of the difference of 90 and 80 ? 
The difference of 80 and 50 is what per cent, of the difference of 
90 and 80 ? 

(b-V)lOO (80 — 50)100 OArt . 
a-b -— 90^80- = 8 °°' AnS - 

If to 2 volumes of 95 per cent, alcohol, 1 volume of 50 per cent, 
alcohol be added, what per cent, alcohol will be the mixture ? The 
sum of 50 and twice 95 is what per cent, of the sum of 2 and 1 ? 

2a -f 6 2X954-50 
Y ^ T =-_^- r — = 80 per cent. Ans. 

In a barrel of apples, the number of sound ones was 60 per 
cent, greater than the number that were damaged. What per 
cent, less was the number that were damaged than the number 
that were sound ? 60 per cent, is what per cent, of the sum of 
100 per cent, and 60 per cent. ? .6 is what rate of 1 -{- .6 ? 

_ a 100 O.a 1 _ 60 

r -n^~ 1 ~l+a~H^- 1 ~l^ _ r+60-"- 375 ' Or 

3 1\ per cent. Ans. 

Since the number of damaged apples was 37^ per cent, less than 
the number that were sound, what per cent, greater was the num- 
ber th? J ^vere sound than the number that were damaged ? 

r = a -r- (1 — a) = 1 -f- (1 — a) — 1 == 60 per cent. Ans. 

Since the number of sound ones was 60 per cent, greater than 

the number that were damaged, what per cent, of the whole were 

sound ? 

a 4- a* 14-. a' 100 -f- 60 0rt 
r = — ^ = — l - — , p= J = 80 per cent. Ans. 

1 What per cent, of the whole were damaged ? 

(100 — 60) ~ 2 = 20 per cent. Ans. 



118 PERCENTAGE. 

Since 20 per cent, of the apples were damaged, what per ceri; 
less was the number that were damaged than the number that wei 
sound ? 

1 — 2. a .. 1 100 — 2a 100 



2 — 2. a 2 — 2.a'^ 200 — 2a 200 — 40 

37J per cent. An 

What per cent, greater was the number that were sound tha 
the number that were damaged ? 

r = 2 — (1 -{-2. a) = 2 — 2. a — 1 = 60 per cent. Ans. 

Since 80 per cent, of the whole were sound, what per cent, les' 
was the number that were damaged than the number that wer 
sound? 

•2. a— 1 . 1 2X-80 — 1 

r = = 1 = — ^ = 37i per cent. Ani 

2. a 2. a 2X-80 2r 

Since the number of damaged ones was 3 1\ per cent, less thai 
the number that were sound, what per cent, of the whole wer^ 
sound ? 

1 100 100 

r== 2=2^' ^ = 2-^2a= 2-2 X 37.5 = 8 °P erCeDt - ^1 

Since 80 per cent, of the whole were sound, what per cent 
greater was the number that were sound than the number tha 
were damaged ? 

r = ~' a = 2.a — 1 = 2 X -80 — 1 = 60 per cent. Ans. 
2 

Lost 20 per cent, of a cargo of coal by jettison, and 5 per cent 
of the remainder by screening, what per cent, of the coal wai 
saved ? 

a — V=d' l r = (l—r>)(l—r")=(l—.20)-yL — .2V 
dl — b" = d" } X -05 = (1 — .20)(1 — .05) = 76 per cent. Ans] 
d" — b"' = d f ", &c. 

Yesterday drew 12 per cent, of my balance of $4,273 in the 
bank, and deposited $1,000; and to-day have drawn 31|- per cent! 
of the balance left over, or as it stood last night. What per cent, o 
the sum of the first-mentioned balance and deposit of yesterday 
have I drawn ? 

b'A-b" 512 + 1487.575 n „ nnet . 

= 37.9354 -{-per cent. Ans. 



a + m 4273 -J- 1000 



PERCENTAGE. 119 

What per cent, of the said sum is remaining in the bank ? 

b'-\-b"_a + m — V — b" a -\-m—(V-\-b") 

~~ ; — i — i := G2.0646 — 

a-\-m a-\-m a-\-m , A 

' ' ' per cent. Ans. 

What per cent., predicating it upon the first-mentioned balance, 
ve I drawn ? 

b' + b" 512.76 + 1487.576 AaQ1QA A A 

r=-J- — = = 46.8134- per cent. Ans. 

a 4273 v 

What per cent, have I drawn, predicating it upon what I now 

ve in the bank ? 

b' + b" _ b' + b" __ 

,- a __&'_jl m _&" — a -\-m— (&' + &")" 61.1225 + 

percent. Ans. 

What amount of money must I deposit to make good 62-| per 

it. of the aforementioned sum ? 

= r (a + m) + &' + &" — (a-f m) =r (a + m) — d" = 

$22.96. Ans. 

To find the Principal or Basis. 

EXAMPLES. 

Che percentage being 250, and the rate .06, what is the 
icipal? 
| = & + r=100&+.p = 250 + .06 = 25,000 + 6 =4,166f. Ans. 

I JL tax at the rate off of 1 per cent, on the valuation was $27.50. 
Ikat was the valuation ? 

bX 6X100 „,„ rtrtn 

a= ^ * =$3,300. Ans. 

5 

•old 120 barrels of flour, which amounted to 12 per cent, of a 
-ain consignment. The consignment consisted of how many 
rels? 

120 + 0.12 = 1,000. Ans. 

16 bushels is more by 8 per cent., or 8 per cent, more, than what 
aber of bushels ? 8 per cent, more than what number is equal 
16 ? What number, plus 8 per cent, of it, will make 216 ? 

a = s+(l+r) = 216 + 1.08=200. Ans. 

30 lbs. is less by 8 per cent., or 8 per cent, less, than what num- 



120 INTEREST. 

ber of lbs. ? 8 per cent, less than what number is 200 ? What 
number, minus 8 per cent, of it, is equal to 200 ? 

a — d -j- (1 — r) = 200 + (1— .08) = 21 7^. Ans. 

. • . 217^-— 217^X-08 = 200=a — b = d = a(l — r). 

To a quantity of silver, a quantity of copper equal to 20 per 
cent, of the silver is to be added, and the mass is to weigh 22 
ounces. What weight of silver is required ? 

a = s-7- (1 -j- r) — 22+1. 2 = 18£ ounces. Ans. 

What weight of copper is required ? 

s sr 

s — ■= — j — = „— r - = 3# ounces. Ans. 
1 -J- r 1-j-r 3 

To a quantity of copper, a quantity of nickel equal to 62J per 

cent, of the copper, a quantity of zinc equal to 33^ per cent, of the 

copper, and a quantity of lead equal to 5 per cent, of the copper, 

are to be added; and the whole is to weigh 401 pounds. The 

weight of each constituent of the alloy is required. 

s 40£ 



a = 



1 _j_ r _j_ r '_|_ r // ~ ! +.. 62£+. 331-f. 05 

= 20 lbs. of copper, 
b =z 20 r = 1 2\ lbs. of nickel, 
b'= 20 r' = 6§ lbs. of zinc, 
yt _ 20 r" = l lb. of lead. 



Ans. 



INTEREST. 

Universal for any rate per cent. 

T = time in months and decimal parts of a month ; t= time in days ; 
P = principal ; r = rate per cent., expressed decimally; i=interest. 

. = PXTXr _ PX<Xr 
12 365 

12j_365i T _12j _365t __12j_365« 
~"Tr~" tr ' ~Pr' _ Pr* ~~PT _ ~¥T' 

Example. — A promissory note, made April 27, 1864, for 



INTEREST. 121 

$825 T ^- and interest at 6 per cent, matured Oct. 6, 1865 : what 
was the interest ? 

Time from April 27 to Oct. 6 (one of 
the dates always included) = 162 days, 
which, added to the 365 days in the year 
preceding = 527 days. 

Note. — One day's interest at least is gener- 
ally lost by computing the time in years and 
months, or months, instead of days. 



Oct. is 10th month. 
April is 4th month. 

Y. m. d. 

1865 . 10 . 6 

'64 . 4 . 27 



Time= 1.5.9 

825.25 X 17.3 X -06-^-12 = $71.38. Ans. 
825.25 X 527 X -06 -f- 365 = $71.49. Ans. 

To find a constant divisor, k, for any given rate per 

When the time is taken in months, k = 1 2 -J- r. 

When the time is taken in days, k = S65—r; thus, 

P X t 
When the rate is 6 per cent. ^^-= Interest. 

P X t 
When the rate is 7 per cent. = Interest, &c. 

Example. — Eequired the interest on $750 for 93 days, at 7 
per cent. 

750 X 93 -f- 5214 = $13.38. Ans. 

Example. — What is the rate per cent, when $450 gains $94^ 
in 3 years? 

450 : 100 ;: 94.5 : 3x = 7 per cent. Ans. 

94.5 -L- 3 X 450 = .0 7. Ans. 

Example. — In what time will $125 at 6 per cent, gain $18|-? 

• 6 : 100 *: 18.75 : 125 X x = 2^ years. Ans. 

18. 75 -^- 1 25 X -06 = 2£ years. Ans. 

Example. — What principal at 5 per cent, interest will gain 
$16J in 18 months? 

5 : 100 ;: 16.875 1 1.5 X* = $225. Ans. 

16.875 X 12 -f- 18 X-05 = $225. Ans. 
11 



122 COMPOUND INTEREST. 

When partial payments have been made. 

Rule. — Find the amount (sum of the principal and interest) 
up to the time of the first payment, and deduct the payment there- 
from ; then find the interest on the remainder up to the next pay- 
ment, add it to the remainder, or new principal, and from the sum 
subtract the next payment ; and so on for all the payments ; then 
find the amount up to the time of final payment for the final 
amount. 



COMPOUND INTEREST. 

If we calculate the interest on a debt for one year, and then on 
the same debt for another year, and again on the same debt for 
still another year, the sum will be the simple interest on the debt 
for three years. But, on the contrary, if we calculate the interest 
on the debt for one year, and then on the amount (sum of the prin- 
cipal and interest) for the next year, and then on the second 
amount for the third year, the sum of the interest so calculated 
will be the compound interest, or yearly compound interest, on the 
debt for three years ; equal to the simple interest on the debt for 
three years, plus the yearly compound interest on the first year's 
interest for two years, plus the simple interest on the second year's 
interest for one year. So, if we divide the time into shorter 
periods than a year, and proceed for the interest as last suggested, 
the interest will be compound. Thus we have half-yearly com- 
pound interest, or compound interest semi-annually, quarter- 
yearly compound interest, or compound interest quarterly, &c. 

This method of computing interest is predicated upon the 
natural idea, that interest, when it becomes due by stipulation and 
is withheld, commences to draw interest, and continues at use to 
the holder, at the same rate as the principal, until it is paid, like 
other over-due demands ; and that the interest so made matures 
and becomes due as often, and at the same periods, as that on the 
principal. 

It will be perceived by the foregoing that the working-time in 
compound interest is the interval between the stipulated payments 
of the interest, or between one stipulated payment of the interest 
and that of another ; and that the working-rate is pro rata to the 
rate per annum. 

Thus the amount of $100 at semi-annual compound interest for 
2 years, at 6 per cent, per annum, is 



COMPOUND INTEREST. 123 

100 X (1.03)* = $112.550881 =$112.55, or 
100. 
.03 

100. 

103~. 
.03 

3.09 
103. 



106.09 
.03 

3.1827 
106.09 

109.2727 
.03 

3.278181 
109.2727 

$112.550881, as before. 

If we let P = principal or debt at interest, 
r = working-rate of interest, 

n = number of intervals into which the whole time is 
divided for the payment of interest, or number of consecutive 
intervals for the payment of interest that have transpired without 
a payment having been made, 

i = compound interest, 
A = P -f- « or amount, then 

^-=(l+r)»;«=A-P. 

Example. — What is the compound interest, or yearly com- 
pound interest, on $100 for 1£ years, at 6 per cent, a year ? 

100X 1.06 X 1-03 = 109.18 — 100 = $9.18. Ans. 

Example. — What is the amount of $560.46, at 7 percent* 
compound interest per year, for 6 years and 57 days ? 

560.46 X (1.07)6 x6 + ' 0y *: 5y > \ = $850.29. Ans, 
\ 365 / 



124 COMPOUND INTEREST. 

Example. — The principalis $250, the rate 8 per cent, a year, 
the time 2 years, and the interest compound per quarter year: 
required the amount. 

250x(l.— )" =$292.91. Ans. 



(^) 8 = 



When Partial Payments have been made. 

Rule. — Find the amount up to the first payment, and deduct 
the payment therefrom ; then find the amount up to the next pay- 
ment, and therefrom deduct that payment ; and so on for all the 
payments ; then find the amount up to the time of final payment, 
lor the final amount. 

Example. — A note of hand for $500 and interest from date, 
at 6 per cent, a year, has been paid in part as follows ; viz., two 
years and four months from the date of the note, by an indorse- 
ment of $50 ; and three years from that indorsement, by an in- 
dorsement of $150. It is now eight months since the last payment 
was made, and the demand is to be settled in full : required the 
amount at the present time, interest being compound per year. 

500 X (1.06) 2 X 1.02 — 50 = 523.036 

(1.06) 3 

622.944 ' 
150 



472.944 
1.04 



$491.86. Ans. 

The following table shows (1 -j- r) raised to all the integer 
powers from 1 to 12 inclusive ; r being taken at 4, 5, 6, 7, 8, and 10 
per cent. If the numbers in the column headed years are taken 
to represent years, then 4 per cent., 5 per cent., &c, at the head 
of the columns of powers, will stand for per cent, per annum : if 
they are taken to represent half-years, then 4 percent., 5 percent., 
&c, will stand for per cent, per half-year, &c. The quantities in 
the columns are powers of (1 -|-r), of which the numbers referred 
to and standing opposite, respectively, are the exponents. Thus, 
1.26248, in the 6 per cent, column, and against 4 in the column 
marked years, = (1.06) 4 ; and so with the others. The powers 
or quantities in the columns are co-efficients in the calculations. 



COMPOUND INTEREST. 



125 



Years. 


4 per cent. 


5 per cent. 


6 per cent. 


7 per cent. 


8 per cent. 


10 percent. 


1 


1.04 


1.05 


1.06 


1.07 


1.08 


1.10 


2 


1.0816 


1.1025 


1.1236 


1.1449 


1.1664 


1.21 


3 


1.12486 


1.15762 


1.19102 


1.22504 


1.25971 


1.331 


4 


1.16986 


1.21551 


1.26248 


1.3108 


1.36049 


1.4641 


5 


1.21665 


1.27628 


1.33823 


1.40255 


1.46933 


1.61051 


6 


1.26532 


1.3401 


1.41852 


1.50073 


1.58687 


1.77156 


7 


1.31593 


1.4071 


1.50363 


1.60578 


1.71382 


1.94872 


8 


1.36857 


1.47746 


1.59385 


1.71819 


1.85093 


2.14359 


9 


1.42331 


1.55133 


1.68948 


1.83846 


1.999 


2.35795 


10 


1.48024 


1.62889 


1.79085 


1.96715 


2.15892 


2.59374 


11 


1.53945 


1.71034 


1.8983 


2.10485 


2.33164 


2.85312 


12 


1.60103 


1.79586 


2.0122 


2.25219 


2.51817 


3.13843 



Note. — If a co-efficient is wanted for a greater number of years or intervals 
of time than is given in the table, square the tabular co-efficient opposite half 
that number of intervals, or cube the tabular co-efficient opposite one-third 
that number of intervals, &c, for the co-efficient required. Thus, 



1.999 2 =1, 



= 1.08 12 X 1.08 c= 1.0818 = 3.996, 



the co-efficient for 18 years or intervals at 8 per cent, per interval, &c. 

If the compound interest alone is sought on a given principal, subtract 1 
from the tabular power corresponding to the time and rate, and multiply the 
remainder by the given principal ; the product will be the compound interest. 
Thus (1.26532 — 1)X 100 = $26,532, the yearly compound interest, at 4 per 
cent, per annum, on §100 for 6 years, or the half-yearly compound interest, at 
8 per cent, per annum, on $100 for 3 years, or the half-yearly compound inter- 
est, at 4 per cent, per half year, on $100 for 6 half-years. 

Example. — What is the amount of $125.54, at 5 per cent, 
compound interest, for 7 years, 21 days ? 

21 X .05 
1 -\ r^ — = 1.00288, the co-efficient for the odd days; and, 

turning to the 5 per cent, column in the table, we find against 7, in 
the column of years, 1 .4071, the co-efficient for 7 years : then 

125.54X1-4071X1.00288 = $178.20. Ans. 

Example. — In what time, at 7 per cent, compound interest 
per annum, will $1000 gain $462? A-^-P = (1 + r) n : then 
1462 -^-1000= 1 .462, the co-efficient demanded. Turning now 
to the 7 per cent, column in the table, we find the nearest less 
co-efficient there (there being none that exactly corresponds) to 

be that for 5 years : viz., 1.40255. And | — '- 1 1 -f- .07 = 

J ' \1. 40255 / 

.60553, the fraction of a year over 5 years to the answer. 

.60553 X 365 = 221 days: 5 years, 221 days. Ans. 
11* 



126 



COMPOUND INTEREST. 



The following table is of the same nature as the preceding, 
and is applicable when the interest becomes due at regular inter- 
vals short of a year, or when the working-rate in compound inter- 
est is less than 4 per cent. 

The«quantities in the If per cent, column apply to quarter-yearly 
compound interest when the rate is 7 per cent, a year ; and those 
in the 1\ per cent, column, to quarterly compound interest when 
the rate is 5 per cent, a year ; also the former are applicable to 
monthly compound interest at 21 per cent, per annum, and the 
latter to monthly compound interest at 1 5 per cent, per annum ; 
and so relatively, throughout the table. 





a 

o 


a 

a 


a 

V 


"5 


"2 


a 

a 


a 


a 






u 










■n 




« 




£ 


a 


<u 


a 


IB 


a 


a 


a 




a 


£ 


a 


S? 


a 


& 


£ 


3 


a 


^ 


1 


1.035 


1.03 


1.025 ,1.02 


1.0175 


1.015 


1.0125 


1.01 


1.005 


2 


1.07123 


1.0609 


1.05063:1.0404 


1.03531 


1.03023 


1.02516 


1.0201 


1.01003 


3 


1.10872 


1.09273 


1.0768911.06121 


1.05342 


1.04568 


1.03797 


1.0303 


1.01508 


4 


1.14752 


1.12551 


1.103811.08243 


1.07186 


1.06136 


1.050951.0406 


1.02015 


5 


1.18769 


1.15927 


1.13141 1.10408 


1.09062 


1.07728 


1.06408 


1.05101 


1.02525 


6 


1.22925 


1.194051.15969:1.12616 


1.1077 


1.09344 


1.0774 


1.06152 


1.03038 


7 


1.27228 


1.22987 


1.1886911.14869 


1.12709 


1.10984 


1.09087 


1.07214 


1.03553 


8 


1.31681 


1.26677 


1.2184 1.17166 


1.14681 


1.12649 


1.10451 


1.08286 


1.04071 


9 


1.3629 


1.30477 


1.24886 1.19509 


1.16688 


1.14339 


1.11831 


1.09369 


1.04591 


10 


1.4106 


1.34392 


1.28008J1.21899 


1.1873 


1.16054 


1.13229 


1.10462 


1.05114 


11 


1.45997 


1.38423 


1.3120911.24337 


1.20808 


1.17795 


1.14645 1.11567 


1.0564 


12 


1.51107 1.42576 


1.3448911.26824 


1.22922 


1.19562 


1.16078 l 1.12683 


1.06168 



Example. — What is the amount of $750 for 4 years and 40 
days, allowing half-yearly compound interest, at 7 per cent, a year ? 

In this case, the working-rate for the full periods of time is 3£ per 
cent., and there are 8 such full periods ; then, seeking the co-efficient 
in the 3^ per cent, column, we find against 8, in the column of 



times, the quantity or co-efficient 1.31681 ; and 1 -f- 
1.00767 : therefore 



40 X-07 
~365^ 



750 X 1.31681 X 1.00767 = $995.18. Ans. 

Example. — What is the amount of $1000 at compound inter- 
est per quarter-year, at 1£ per cent, per quarter-year, for 4^ years ? 

1000 X 1.12649 2 X 1.015 = $1288.01. Ans. 



BANK INTEREST. 127 



BANK INTEREST OR BANK DISCOUNT. 

A bank loans money on a promissory note made payable with- 
out interest at a future period. The operation is called discounting 
the note at bank, and is as follows : The bank takes the note, finds 
the interest on it for three days more time than by its own tenor it 
has to run, subtracts it from the principal, and hands the balance, 
called the avails of the note, in its own bills, to the party soliciting 
the loan, or offering the note for discount, as it is called ; whereby 
the note becomes the property of the bank, and the maker and 
indorsers are held for its payment when it matures. 

The three days mentioned are called days of grace, and the 
note does not become due to the bank until three days after it 
becomes due by its own tenor. These proceedings are sanctioned 
by usage, and protected by law. 

Bank interest, then, is bank discount, and bank discount is bank 
interest. But bank discount is not discount, nor is it what is called 
legal interest on the money loaned. It is the interest on the money 
loaned, plus the interest on the interest of the loan, plus the inter- 
est on the difference of the sum taken and the interest on the loan 
for the time of the loan ! A kind of interest more onerous, if any. 
description of interest be onerous, than compound interest, rate 
for rate and time for time, as may be readily perceived. 

Let P = principal or face of the note. 

r = working-rate of the interest for the time of the loan. 
a = avails of the note or sum borrowed. 
i = bank interest. 
t r= time of the loan. 

R : r : : T : t. R being the rate per cent, per annum, and T 
one year. 

P=a^(l— r). a = V — Pr. i=Pr. r = (P — fl) -f- P. 

If we let n represent the time of the note in months, 
_Rn , 3R 

r — ~y% ~i ogi* But it is the practice with many banks to count 

the days of grace as so many 360ths of a year. 

Putting d to represent the time of the note in days, 

Rt/-f-3R 

r:= , true time and rate. 

365 

With some banks, it is the practice, in calculating interest, to take 
the time, when it does not exceed 93 days, as so many 360ths of a 
year. 

A note having 3 months to run from Aug. 10, for instance, will 



128 



BANK INTEREST. 



fall due Nov. 10-13; but one having 90 days to run from Aug. 
10 will fall Nov. 8-11. The time including grace of the former 
*is 3 mo. 3 ds., and that of the latter 3 mo. 2 ds., mean time. Never- 
theless, the former embraces 95 days, or one day more than mean 
time, and the latter but 93 days. 

The following table shows 1 — r, mean time, for the intervals of 
time set down in the left-hand column ; B. being taken at 4, 5, 6, 7, 
and 8 per cent, per annum, as set down at the top of the columns. 



Time. 




4 


5 


6 


7 


8 


mo. 


ds. 


per cent. 


per cent. 


per cent. 


per cent. 


per cent. 


1 


3 


.996333 


.995417 


.9945 


.993583 


.99266 7 


2 


3 


.993 


.99125 


.9895 


.98775 


.986 


3 


3 


.989667 


.987083 


.9845 


.981917 


.979333 


4 


3 


.986333 


.982917 


.9795 


.976083 


.972667 


5 


3 


.983 


.97875 


.9745 


.97025 


.966 


6 


3 


.979667 


.974583 


.9695 


.964417 


.959333 


7 


3 


.976333 


.970417 


.9645 


.958583 


.952667 


8 


3 


.973 


.96625 


.9595 


.95275 


.946 


9 


3 


.969667 


.962083 


.9545 


.946917 


.939333 


10 


3 


.966333 


.957917 


.9495 


.941083 


.932667 


11 


3 


.963 


.95375 


.9445 


.93525 


.926 


12 


3 


.959667 


.949583 


.9395 


.929417 


.919333 



Putting k to represent the tabular quantity 1 — r, 

a— Pfc, P— a-i- *, t = P — a = P — PL 

Example. — What will be the avails of a note for Si, 250 
payable in 4 months if discounted at a bank, interest being 7 per 
cent, a year ? 

The tabular constant 1 — r, in the 7 per cent, column, against 4 
months and 3 days in the time column, is .976083, and 
$1,250 X .976083 = $1,220.10. Ans. 
Example. — For what sum must I make a note having 6 months 
to run, in order that the avails at bank, if discounted on the day 
of the date of the note, may amount to $956.38, interest being 6 
per cent, per annum ? 

By the table, $956.38 -f- .9695 = $986.47. Ans. 
Example. — What is the rate of bank interest when the nomi- 
nal or legal rate is 7 per cent. ? 

.07 -J- (1 — .07) = .07527 = 1\ -f- T f £_ per cent. Ans. 

Note. — A note having 5 months to run from Feb. 1 will fall due July 
1-4; and the time, including grace, is 5 mo. 3 ds.= 155 days, mean time. 
But the time in days from Feb. 1 to July 4, when February has but 28 days, 
is 153 days only, or 2 days short of mean time. See Sec. B., p. 3. 



DISCOUNT. COMPOUND DISCOUNT. 129 



DISCOUNT. 

Discount is a deduction of the interest on the present worth or 
availability of a debt not yet due, in consideration of its present 
payment. The, principal is the present nominal value of the debt, 
interest included, if any interest has accrued. The time is the 
interval from the present to the date at which the debt will become 
due. The rate is the legal rate of interest, if no other rate is speci- 
fied ; and the present worth is that sum of money, which, if put at 
interest at the same rate and for the same time as the discount, will 
amount to the principal. 

Let a represent the principal, d the discount, w the present 
worth, and i the interest on one dollar for the time and at the rate 
of the discount. 

w = a -J- (1 -j- i) = a — d. dz=ai~(l-\- i) = a — w. 
a = d (1 -}- i) -j-i— d-\-w. 

Example. — Required the discount on $250 for 8 months at 6 
per cent. 

The interest on $1 for 8 months at 6 per cent, is .04 of a dollar, 
or 4 cts. ; and 

250 X .04-f- (l-[- .04) = $9.6154. Ans. 

Example. — Required the present worth of $1272.62 due 247 
days hence, discount 7 per cent. 

The interest on $1 for 247 days at 7 per cent. = 247 X -07 -4- 365 
= 0.04737, and 

1272.62-4- 1.04737 =$1215.06. Ans. 

Kote. — " Talcing off, in common parlance, a certain per centum from 
the face of a demand, is equal to deducting the interest, at that rate per cent- 
turn, on the present worth for 1 year, plus the interest on the interest of the 
present worth, at the same rate per centum for 1 year. Sse Sec. B., p. 1. 



COMPOUND DISCOUNT. 

Compound Discount is to compound interest what simple dis- 
count is to simple interest. In both cases of discount, the differ- 
ence between the principal and the discount is that sum of money, 
which, if put at interest for the same length of time, at the same 
rate, and in the same general manner as the discount, will amount 
to the principal. 

Rule. — Add 1 to the rate per cent, of the discount for the 



130 COMPOUND DISCOUNT. 

working-time, and raise the sum to a power corresponding with the 
number of working-times ; divide the principal by the power, and 
the quotient will be the present worth ; subtract the present worth 
from the principal, and the remainder will be the compound 
discount. 

Note. —The tables of the powers of 1 + r, applicable to compound in- 
terest, are equally applicable to compound discount. 

Example. — Required the present worth of a debt of $250, 
allowing yearly compound discount, at 7 per cent, a year, for 
3 years 84 days. 

07 X 84 

1 _|_ * __ = 1.01611, the working-rate for the 84 days, and 

365 

250 -f- (1.07 3 X 1.01611) = $200.84. Ans. 

Example. — What is the present worth of a debt of $150.25, 
due 3 years, 3 months, and 10 days hence, without interest, allow- 
ing compound discount per quarter-year, at 1£ per cent, per quar- 
ter-year ? 

150.25 -h ( 1.015 13 X 1. '° 6 X 10 ) = Ans. 
\ 365/ 

By table, 150.25-^(1.19562 X 1-015 X 1.00164) = 

$123.61. Ans. 

Note. — What is here denominated the debt, or principal, represents the 
debt at the close of the time of the discount; that is, if the debt be on in- 
terest, the iuterest must be included in what is here called the debt, or 
principal. 



PROFIT AND LOSS. 

The term "Profit and Loss," as intimated in treating of 
Percentage, relates to the positive and negative interests in 
business, and embraces the idea of both. 

Both profit and loss are absolute quantities, and are expressed by 
the difference of the cost price and selling price that limit them. 
They are usually, however, estimated by percentage, predicated 
upon the first-mentioned price or prime cost. 

When the selling price is greater than the cost price, or when 
the money obtained by the disposal of property exceeds what the 
property cost, the difference is positive, and denotes increase, 
profit, or gain. Conversely, when the cost price is greater than the 
selling price, or when property is disposed of for less money than 
it cost, the difference is negative, and denotes decrease, loss, or 



PROFIT AND LOSS. 131 

waste. So, the difference of the two prices, divided by the cost 
price, expresses the rate of gain on the cost when the selling price 
is the greater, — expresses the rate of loss on the cost when the 
cost price is the greater. 

Let c represent the cost price, purchase price, par value, or sum 
of money paid for the property ; s, the selling price, trade price, 
premium price, or sum of money received in exchange for the 
property ; r, the rate of the profit or loss ; p, the rate per cent, of 
the profit or loss. 

To find the rate or rate per cent, of the profit or loss. 

s ^, c (s ~ c} 100 ,. . . -.„ 

r = . p — ^ ^ . Moreover, when the difference is 



s s 
positive, ?'= 1 ; and, when it is negative, r — 1 . 

Example. — Paid $4 for an article, and sold it for $5. What 

per cent. was. gained ? 5 is more than 4 by what per cent, of 4 ? 

The difference of 5 and 4 is what per cent, of 4 ? 5 — 4 = $1, 

5 ~ 4 
gained; and :=:.25^r£— 1. 25 per cent. Ans. 

Example. — Paid So for an article, and sold it for $4. What 

per cent, was lost ? 4 is less than 5 by what per cent, of 5 ? The 

difference of 4 and 5 is what per cent, of 5? 4 — 5 =z — 1 = $1, 

5 ~ 4 
lost ; and — - — — .20 = 1 — \. 20 per cent. Ans. 
o 

Example. — A whistle that cost 3 cents was sold for 20 cents ! 
The profit was how much per cent ? (20 ~ 3) -f- 3 = 5§ or 566§ 
per cent. Ans. 

Example. — A fop paid $10 for a well-made and well-fitting 
pair of boots for his own wear, that were worth what they cost him ; 
but, being told that they were unfashionably large, sold them for 
$4. His vanity cost him what per cent, of the purchase price ? 
1 — T 4 ? =z .6 or 60 per cent. Ans. 

To find a price long a given per cent, of the cost, or to find a sell- 
ing price that shall be the sum of the cost price and a given per 
cent, of it. 

s = c-f-cr = c(l-L-r) = c (100 -|- p) -f- 100. 

Example. — At what price must I sell an article that cost 
S2.35 to gain 25 per cent. ? 2.35, more 25 per cent, of it, is how 
much ? The sum of $2.35 and 25 per cent, of it is how much ? 
2.35 -4- 2.35 X -25 = 2.35 X 1-25 = $2.93f. Ans. 



132 EQUATION OF PAYMENTS. 

To find a price short a given per cent, of the cost, or to find a sell- 
ing price that shall be the difference of the cost price and a given 
per cent, of it. 

s—c — cr—c (1 — r)=zc (100 — p) -J- 100. 

Example. — I have a damaged article of merchandise that cost 
$2.75, and I wish to mark it for sale at 30 per cent, below cost. 
At what price shall I mark it ? 2.75 less 30 per cent, of it is how 
much? The difference of $2.75 and 30 per cent, of it is how 
much? 2.75 (1— .30)^2.75 X -7 = $1,925. Ans. 

To find the cost price when the selling price and profit per cent, are 

given. 
s = c-{-cr = c (1+/*). • . c=s-j-(l -f-r)~ 100 s -£-(100 -f-;>). 

Example. — What cost that article whose selling price, $4, is 
long 25 per cent, of the cost ? What price, more 25 per cent, of 
it, is equal to $4 ? $4 is the sum of what price and* 25 per cent, 
of it? 400-^-125 = $3.20. Ans. 

To find the cost price when the selling price and loss per cent, are 

given. 

s = c — cr — c(l — r) .-. c=ls-±- (1 — r) = 100s -^- (100 — p) 

Example. — What cost that article whose selling price, $375, 
is short 7 per cent, of the cost ? What price less 7 per cent, of it 
is equal to $375 ? $375 is the difference of what price and 7 per 
cent, of it ? 
375 -7- (1 — .07) = 375 -£- .93 = 375 X 100 -"- (100 — 7) = 

$403,226. Ans. 



EQUATION OF PAYMENTS. 

, The Equation of Payments, or Averaging of Accounts, as 
it is more frequently called, practically consists in finding the 
common time of maturity of two or more debts due at different 
times, and is either special or general ; special when it is made in 
regard to a given interchangeable rate of interest and discount, in 
which the magnitude of the rate slightly affects the time, since 
discount consumes more time per dollar, rate for rate, than inter- 
est ; and general, when it is made disrespectful of rate, or com- 
mon in the greatest possible degree to all rates. 

Rule, for common purposes. — Multiply each debt by the num- 
ber of days between its own date of maturity and that of the debt 
earliest due, and divide the sum of the products by the sum of the 
debts ; the quotient will express the common time in days subse- 
quent to the leading date. 



EQUATION OF PAYMENTS. 3 

The following exhibits the face of an account in the edger, and 
♦he time (date) at which it averages due is required. 

1960, April 10 $250.26 — 6 mo. Due Oct. 10. 

" June 25 320.56 — 6 " " Dec. 25. 

" Julv 10 50.62 — 3 " " Oct. 10. 

" Aug. 1 210.84 — 4 " " Dec. 1. 

" 18 73.40 — 5 " " Jan. 18. 

" Oct. 15 100. —cash " Oct. 15. 

Example. — Practical method of stating and working. 
1860. Due Oct. 10, $301 

" " Dec.25, 321 X 76 =■ 24396. 

" " " 1, 211 X 52= 10972. 

" " Jan. 18, 73 X 100 = 7300. 

" " Oct. 15, 100 X 5 = 500. 

1006" ) 43168 ( 43 days, = Nov. 22, 1860. 

Arts. 
COMPOUND AVERAGE. 

Compound Average consists in rinding the time at which the bal- 
ance of an account or demand averages due, whose sides — the debit 
and the credit — average due at different dates. 

Rule. — Multiply the less sum or side by the difference in days 
between the two dates — that at which the debit side averages due 
and that at which the credit side averages due — and divide the prod- 
uct by the difference of the sums or sides ; the quotient will be the 
number of days that one of the dates must be set back, or the other 
forward, to mark the time sought ; for which last, 

special rule. 
Earlier date with larger sum, set back from earlier. 
Later date with larger sum, set forward from later. 

Example. — The debit side of an account in the ledger foots up 
$400, and averages due Oct. 12, 1860 ; the credit side of the same 
account foots $300, and averages due Nov. 16, 1860. At what d-ate 
does the balance or difference between the two sides average due ? 

400 300 

300 35 

"lOO ) 10500 ( 105 days earlier than Oct. 12, = June 29, 1860. Ans. 

Example. — The debit side of an obligation foots $250, and aver- 
ages due May 17, 1860; the credit side of the same obligation foots 
$175, and averages due May 1, 1860. At what date does the differ- 
ence of the sides average due 1 
250 175 . • 

175 16 

75 ) 280Q ( 37J days later than May 17, = June 23, 1860. Ans. 
12 



134 GENERAL AVERAGE. 



GENERAL AVERAGE. 

It is the established usage that whatever of either of the three 
commercial interests — the ship, the cargo, or the freight — is 
voluntarily sacrificed or destroyed for the general good, or with 
the view of saving the most that may be saved when all is in immi- 
nent danger of being lost, is matter of general loss to the respec- 
tive interests, and not more especially to the interest voluntarily 
abandoned than to the others. So, too, the losses and damages inci- 
dent to the voluntary sacrifice, and collateral therewith, together 
with the expenditures which the master has been compelled to 
make for the general good, in consequence of disaster, are matters 
of general average, or are to be contributed for, pro rata, by the 
several interests. 

The contributory interests are the ship, the cargo, and the 
freight, at their net values, independent of charges, premiums 
paid for insurance, &c. 

The contributory value of the ship, generally, is her value at the 
port of departure at the time of leaving, less the premium paid for 
her insurance. 

The contributory value of the cargo is its net value, in a sound 
*state, at the port of destination, if the voyage be completed ; or its 
invoice value if the voyage be broken up and the cargo returned 
to the port whence it was shipped ; or its market-value at any in- 
termediate port, where of necessity it is discharged and disposed of. 
The value of the goods jettisoned, and to be contributed for, is 
their value after the same manner ; and that value is a part of the 
contributory value of the cargo, as well as a matter of general 
average. 

The contributory value of the freight, generally, is the gross 
amount or amount per freight-list, less one-third part thereof, in 
most of the States ; but, in the State of New York, one-half thereof, 
for seamen's wages and other expenses. The loss of freight by 
jettison, when any freight is earned, is matter of general average. 
If the cargo is transshipped on board another vessel, and in that 
way sent to the port of destination, the contributory value of the 
freight is the gross amount, less the sum paid the other vessel. 

The voluntary damage to the ship, with a view to the general 
good, — such as throwing over her furniture, destroying her equip- 
ments, cutting away her masts, breaking up her decks to get at the 
cargo for the purpose of throwing it over, &c, — is contributed for 
at two-thirds the cost of repairing and restoring ; the new articles 
being supposed one-half better, or worth one-half more, than the 
old. 



GENERAL AVERAGE. 135 

If we let V = contributory value of the vessel, 
C = contributory value of the cargo, 
F = contributory value of the freight, 
d = aggregate amount of losses to be averaged, then 
d -J- (V -{- C -j- F) — r, the per cent, of each interest that each 
must contribute, and 

VX?' — Vessel's share of the loss, 
C X r ' == - Cargo's share of the loss, 
Fx r = Freight's share of the loss. 
When a contributory interest's share of the loss is to be distrib- 
uted among the several owners of that interest, the same pro rata 
method is to be observed : thus 

A X r = sum A must contribute, 
B X r = sum B must contribute, 
D X r = sum D must contribute ; 
A, B, and D being A's, B's, and D's respective shares in that 
interest. 



136 ASSESSMENT OF TAXES. INSURANCE. 



ASSESSMENT OF TAXES. 

G = amount of taxable property, real and personal, as per 
grand list. 

A = amount of money to be raised, including the whole poll-tax. 
. T = amount of money to be raised on property alone. 
n — number of ratable polls. 
h — poll-tax per head. 

r = rate per cent, to be raised on taxable property. 
P = an individual's taxable property, as per grand list. 
b = P's poll-tax. 

T = A — hn. r = T -f- G. P r + & = Ps tax, including poll. 



INSURANCE. 

Insurance is a written contract of indemnity, called the policy, 
by which one party (the insurer or underwriter) engages, for a 
stipulated sum, called the premium (usually a per cent, on the 
value of the property insured), to insure another against a risk or 
loss to which he is exposed. 

Let P = Principal, or amount insured on, 
r = rate per cent, of insurance, 
a = premium for insurance. 

a = Pa r = a — P. P = a — r. 

Example. — What is the premium for insuring on $4500 at l£ 
per cent. ? 

4500 X-015 = $67.50. Ans. 



LIFE-INSURANCE. 

Life-insurance is predicated upon the even chance in years, 
called the expectation of life, that an individual in general health 
at any given age appears by the rates of mortality to have of living 
beyond that age. 

The Carlisle Tables of Expectation, column C in the following 
tables, are used almost or quite exclusively in England, and by 
some insurance-companies in the United States ; while those by 
Dr. Wigglesworth, column W, computed with special reference to 
the rates of mortality in this country, are used by others. 

The Supreme Court of Massachusetts has adopted the Wiggles- 



LIFE INSURANCE. 



137 



■worth rates of expectation in estimating the value of life-annuities 
and life-estates. 

TABLE 

Of Ages and Expectations from Birth to 103 Years. 



Age. 


c. 


w. 


Age. 


c. 


W. 


Age. 
52 


c. 


W. 


Age. 


c. 


w. 
6.59 





38.72 


28.15 


26 


37.14 


31.93 


19.68 


20.05 


78 


6.12 


1 


44.68 


36.78 


27 


36.41 


31.50 


53 


18.97 


19.46 


79 


5.80 


6.21 


2 


47.55 


38.74 


28 


35.69 


31.08 


54 


18.28 


18/92 


80 


5.51 


5.85 


3 


49.82 


40.01 


29 


35.00 


30.66 


55 


17.58 


18.35 


81 


5.21 


5.50 


4 


50.76 


40.73 


30 


34.34 


30.25 


56 


16.89 


17.78 


82 


4.93 


5.16 


5 


51.25 


40.88 


31 


33.68 


29.83 


57 


16.21 


17.20 


83 


4.65 


4.87 


6 


51.17 


40.69 


32 


33.03 


29.43 


58 


15.55 


16.63 


84 


4.39 


4.66 


7 


50.80 


40.47 


33 


32.36 


29.02 


59 


14.92 


16.04 


85 


4.12 


4.57 


8 


50.24 


40.14 


34 


31.68 


28.62 


60 


14.34 


15.45 


86 


3.90 


4.21 


9 


49.57 


39.72 


35 


31.00 


28.22 


61 


13.82 


14.86 


87 


3.71 


3.90 


10 


48.82 


39.23 


36 


30.32 


27.78 


62 


13.31 


14.26 


88 


3.59 


3.67 


11 


48.04 


38.64 


37 


29.64 


27.34 


63 


12.81 


13.66 


89 


3.47 


3.56 


12 


47.27 


38.02 


38 


28.96 


26.91 


64 


12.30 


13.05 


90 


3.28 


3.43 


13 


46.51 


37.41 


39 


28.28 


26.47 


65 


11.79 


12.43 


91 


3.26 


3.32 


14 


45.75 


36.79 


40 


27.61 


26.04 


66 


11.27 


11.96 


92 


3.37 


3.12 


15 


45.00 


36.17 


41 


26.97 


25.61 


67 


10.75 


11.48 


93 


3.48 


2.40 


16 


44.27 


35.76 


42 


26.34 


25.19 


68 


10.23 


11.01 


94 


3.53 


1.98 


17 


43.57 


35.37 


43 


25.71 


24.77 


69 


9.70 


10.50 


95 


3.53 


1.62 


18 


42.87 


34.98 


44 


25.09 


24.35 


70 


9.18 


10.06 


96 


3.46 




19 


42.17 


34.59 


45 


24.46 


23.92 


71 


8.65 


9.60 


97 


3.28 




20 


41.46 


34.22 


46 


23.82 


23.37 


72 


8.16 


9.14 


98 


3.07 




21 


40.75 


33.84 


47 


23.17 


22.83 


73 


7.72 


8.69 


99 


2.77 




22 


40.04 


33.46 


48 


22.50 


22.27 


74 


7.33 


8.25 


100 


2.28 




23 


39.31 


33.08 


49 


21.81 


21.72 


75 


7.01 


7.83 


101 


1.79 




24 


38.59 


32.70 


50 


21.11 


21.17 


76 


6.69 


7.40 


102 


1.30 




25 


37.86 


32.33 


51 


20.39 


20.61 


77 


6.40 


6.99 


103 


0.83 





Thus, by the tables, a man in general good health at 21 years of 
age has an even chance, by the Carlisle rate of mortality, of living 
40| years longer; by the Wigglesworth rate, of living 33 T 8 -^ 
years longer. So a man in general good health, at 60 years of 
age, has, by the Carlisle rate,' an even chance of living 14.34 years 
longer; by the Wigglesworth rate, an even chance of living 15.45 
years longer, etc. 
12* 



138 



FELLOWSHIP. 



FELLOWSHIP. 



Fellowship calls for the distribution of a given effect to each 
of the several causes associated in its production, proportional to their 
respective magnitudes one with another. 

It is a rule, therefore, adapted to the use of partners associated in 
business, in achieving a pro rata distribution among themselves as indi- 
viduals, of the profits or losses pertaining to the company. 

Rule. — Multiply each partner's investment or share of the capital 
stock, by the whole gain or loss, and divide the product by the sum 
of all the shares, or gross capital. 

Example. — Three men, A, B, and C, enter into partnership. A 
invests $500, B $700, and C $300. They trade and gain $400. 
What is each partner's share of the profits? 

A, $500 500 X 400 -f- 1500 == $133. 33J = A's share. 

B, 700 700 X 400 -:- 1500 = 186.66f = B's " 

C, 300 300 X 400 -r- 1500 == 80.00 = C's " 

$1500 = gross capital. $400.00 Proof. 

Example. — D's investment of $600 has been employed eight 
months; E's, of $500, five months; and F's, of $300, five months , 
the profits of the company are $500, and are to be divided pro rata 
among the partners. What is each partner's share 1 

D, $600 X 8 = 4800 X 500 -j- 8800 = $272.73, D's share. 

E, 500 X 5 = 2500 X 500 -4- 8800 = 142.05, E's " 

F, 300 X 5 = 1500 X 500 -j- 8800 = 85.22, F's " 

8800 $500. Proof. 

Example. — Of $120 distributed, there were given to A, \ ; to B, 
\ ; to C, ^ ; and to D, ^, and there was nothing remaining. 
What sum did each receive ? 

\ of 120 = 40 X 120 -r- 114 = $42 T 2 F = A's share. 
J of 120 = 30 X 120 -r- 114 = 31|4 = B's " 
£ of 120 = 24 X 120 -f-**114 = 25 T 5 9 = C's " 
| of 120 = 20 X 120 -j- 114 = 21 T V = D's " 
"Tli $120. Proof. 

Example. — Divide the number 180 into 3 parts, which shall 
be to each other as 1, i, i. 

| of 180 = 90 X 180 -^ 195 = 83.08 
a of 180 — 60 X 180 -f- 195 = 55.38 
\ of 180 = 45 X 180 -^ 195 = 41.54 

195 180.00 Proof. 



ALLIGATION. 139 

Example. — $400 are to be divided between A, B, and C, in 

the ratio of | to A, |- to B, and | to C; how much will each 
receive ? 

X of 400 = 200, and 200 X 400 -f- 500 =$160 = A's share. 
I of 400 = 200, and 200 X 400 -f- 500 = 160 = B's share. 
| of 400 = 100, and 1 00 X 400 -^- 500 = 80 = C's share. 

500 $400. Proof. 



ALLIGATION. 

Alligation Medial is a method by which to find the mean price of 
a mixture or compound, consisting of two or more articles or ingre- 
dients, the quantity and price of each being given. 

Rule. — Multiply each quantity by its price, and divide the sum of 
the products by the sum of the quantities ; the quotient will be the 
price per unity of measure of the mixture ; and, having found the 
price of the given quantities as mixed, any quantities of the same 
materials, taken in like proportions, will be at the same price. 

Example. — If 20 lbs. of sugar at 8 cents, 40 lbs. at 7 cents, and 
80 lbs. at 5 cents per pound, be mixed together, what will be the mean 
price, or price per pound, of the mixture? 

20 X 8 = 160 

40 X 7 = 280 

80 X 5 = 400 

140 ) 840 ( 6 cents. Ans. 

The several kinds, then, at their respective prices, taken in the 
proportion of 1 at 8, 2 at 7, and 4 at 5 cts., will form a mixture worth 
6 cts. a pound. 

Example. — If 10 lbs. of nickel are worth $2, and 24 lbs. of copper 
are worth $4£, and 8 lbs. of zinc are worth 40 cts., and 1 lb. of lead is 
worth 5 cts., what are 5 lbs. of pretty good German silver worth? 
(2oo±_4_5 o^+io + siKA = 81 cents. Ans. 

Alligation Alternate is a method by which to find what quantity 
of each of two or more articles or ingredients, whose prices or quali- 
ties are given, must be taken to form a mixture or compound that 
shall be at a given price or of a given quality between the two 
extremes. It also applies to the finding of relative quantities when 
the quantity of one or more of the articles is limited. 

Rule. — Connect the given prices or qualities — a less than the 
given mean with that one or either one that is greater — and to the 
extent that all be thus connected ; then place the difference between 



140 



ALLIGATION. 



each given and the given mean opposite, not the given, or the 
given mean, but the given with which it is alligated ; the num- 
ber standing opposite each price or quality will be the quantity that 
must be taken at that price, or of that quality, to form a mixture or 
compound at the price or of the quality desired. And, being propor- 
tions respectively to each other, they may be taken in ratio greater 
or less, as desired. 

Example. — In what proportions shall I mix teas at 48 cents a 
pound and 54 cents a pound, that the mean price may be 50 cents a 
pound? 

la the proportions 
48n J 4 lbs at 48 cts. > A 
54J \ 2 lbs. at 54 cts. $ Ans ' 
Or, as 2 at 48 to 1 at 54. 



50 



Proof 52X48 + 1X54 
rrooi. "^ 3 x 5 



150. 
150. 



Example. 
cents a pound. 



60 



48" 
54-] 

721 



-In what proportions shall I mix teas at 48, 54, and 
that the mixture may average 60 cents a pound 1 



12, 
12, 

12 + 6, 



12 at 48 
12 at 54 

18 at 



48 ) ( 2 at 48 ) 

54 ) = { 2 at 54 } Arts. 

72 ) ( 3 at 72 ) 



Example. — A wine dealer has received an order for a quantity of 
wine at 50 cts. a gallon. He has none ready manufactured at that 
price. He has it at 40 cts., at 56 cts., and at 80 cents a gallon, and 
he has water that cost him nothing. He wishes to fill the order 
with a mixture composed of the four materials — the water and the 
three different priced wines. In what proportions must he mix them, 
that the mean or average price may be 50 cents 1 




Or, 50 



Ans. 

30 = 30" 

6 + 30 = 30 

10 =10 

50+10 = 60 



= 96 gals. 
Ans. 



= 136 gals. 
Ans. 




00 



Or, 50- 



404 6 + 30 
56 J 50+10 
80-1 10 



= 176 



= 112 gals. 

If, now, having found the proportions desired, it is wished to limit 
one of the articles in quantity — say the best wine to 8 gallons in the 



INVOLUTION EVOLUTION. 141 

mixture — the pioportions of the remaining articles thereto are found 
thus : — 

Instance, 1st example, — 

i n • q * • Qn ~" oa ( ^nd tne m i xtuTe w iH consist of 

lo ! 8 :! 6 = % ) 8 + 40 + 24 + 4 t = 76 t g allons - 

If, instead, it is desired to mix a given quantity, say 100 gallons, 
and proportioned, say as in first example, the quantity to be taken of 
each is ascertained by the following 

Rule. — As the sum of the relative quantities is to the quantity 
required, so is each relative quantity to the quantity required of it 
respectively. 

The sum of the relative quantities alluded to is 6 -{- 30 -|- 50 -+- 10 
.= 96 ; then, 

96 : 100 :: 6 = 6| 
96 : 100 :: 30=314 
96 : 100 :: 50=52^ 
96 : 100 :: io=io T & 2 



INVOLUTION. 

Involution consists in involving, that is, in multiplying a number 
one or more times into itself. The number so involved is called the 
root, and the product arising from such involution, its power. 

The second power, or square, of the root, is obtained by multiplying 
the root once into itself, as 4 X 4= 16 ; 4 being the root and 16 its 
square. 

The third power, or cube, of a number, is obtained by multiplying 
the number twice into itself, as 4 X 4 X 4 = 64 ; and so on for any 
power whatever. 

When a number is to be involved into itself, a small figure called 
the index or exponent is placed at its right, indicating the number of 
times it is to be so involved, or the power to which it is to be raised. 
Thus, 3 4 = 3 X 3X 3X 3 = 81; and4 3 = 4 X4 X 4 = 64. 



EVOLUTION. 

Evolution is the opposite of Involution. It consists in finding a 
root of a given number, instead of a power of a given root. 

When the root of a number is required or indicated, the number is 
written with the */ before it : and the character or denomination of 
the root, if it be other than the square root, is defined by an index 



142 EVOLUTION. 

figure placed over the sign. When the square root of a number is 
required, the sign (V) is placed before the number, but the index 
(2) is usually omitted. Thus, /\/25, shows that the square root of 
25 is required, or to b» taken ; and 4/25 shows that the cube root 
is required. The operation is usually called extracting the root. 

TO EXTRACT THE SQUARE ROOT. 

Rule — 1. Separate the given number into periods of two figures 
each, by placing a point over the first figure, third, fifth, &c, counting 
from right to left — the root will consist of as many figures as there 
are periods. 

2. Find the greatest square in the left hand period, and place its 
root in the quotient ; subtract the square of the root from the left 
hand period, and to the remainder bring down the next period for a 
dividend. 

3. Multiply the root so far found — the figure in the quotient — by 
2, for a divisor ; see how many times the divisor is contained in the 
dividend, except the right hand figure, and place the result (ihe num- 
ber of times it is contained) in the quotient, to the right of ihe figure 
already there, and also to the right of the divisor ; multiply the divi- 
sor, thus increased, by the last figure in the quotient, and subtract the 
product from the dividend, and to the remainder bring down the next 
period for a dividend. 

4. Multiply the quotient — the root so far found (now consisting of 
two figures) — by 2, as before, and take the product for a divisor ; 
see how many times the divisor is contained in the dividend, except 
the right hand figure, and place the result in the quotient, and to the 
right of the divisor, as before ; multiply the divisor, as it now stands, 
by the figure last placed in the quotient, and subtract the product from 
the dividend, and to the remainder bring down the next period for a 
dividend, as before. 

5. Multiply the quotient (now consisting of 3 figures) by 2, as 
before, and take the product for a divisor, and in all respects proceed 
as when seeking for the last two figures in the quotient. The quo- 
tient, when all the periods have been brought down and divided, will 
be the root sought. 

Note. — 1. If there is a remainder after finding the integer of a root, annex periods of 
ciphers thereto, and proceed as when seeking for the integer. The quotient figures will 
be the decimal portion of the root. 

2. If the given number is a decimal, or consists of a whole number and decimal, point 
off the decimal from left to right, by placing the point over the second, fourth, sixlh, fcc, 
figures therein, and fill the last period, if incomplete, by annexing a cipher. 

3. If the dividend does not contain the divisor, a cipher must be placed in the quotient, 
and also at the right of the divisor, and the next period brought down; then the dividend 
must be divided by the divisor as increased. 

4. If the quotient figure, obtained by dividing by the double of the root, is too large, as 
will sometimes be the case, (see 3d Example) it must be dropped, and a less — one which 
U the true measure — taken in its stead. 



EVOLUTION. 143 



Example. — Required the square root of 123456.432. 

123456.4320 ( 351. 3636+. Ans. 
9 

65 ) 334 
325 



70 . ) 956 
701 



7023 ) 25543 
21069 



70266 ) 447420 
421596 



702723 ) 2582400 
2108169 



7027266 ) 47423100 
42163590 

5259504 

Example. — Required the square root of 10621. Also, of 28561 



10621 ( 103.05-}-. Ans. 
1 



203 ) 00621 
609 
20605 ) 120000 
103025 



16975 



28561 ( 169. Ans. 
1 



26 ) 185 
156 



329 ) 2961 
2961 



TO EXTRACT THE CUBE ROOT. 

Rule — 1. Separate the given number into periods of three figures 
each, by placing a point over the first, fourth, seventh, &c, counting 
from right to left — the root will consist of as many figures as there 
are periods. 

2. Find the greatest cube in the left hand period, and place its root 
in the quotient ; subtract the cube of the root from the left hand pe- 
riod, and to the remainder bring down the next period for a dividend. 

3. Multiply the square of the quotient by 300, for a divisor ; see 
how many times the divisor is contained in the dividend, and place 
the result (except that the remainder is large, diminished by one or 
two units) in the quotient. 

4. Multiply the divisor by the figure last placed in the quotient, 
and to the product add the square of the same figure, multiplied by the 
other figure, or figures, in the quotient, and by 30 ; and add also thereto 



144 



EVOLUTION. 



the cube of the same figure, and take the sum for the subtrahend ; sub- 
tract the subtrahend from the dividend, and to the remainder bring 
down the next period for a dividend, with which proceed as with the 
preceding, so continuing until the whole is completed. 

Note — 1. Decimals must be pointed from left to right, by placing a point over the 
third, sixth, &c, figures in that direction. 

2. If the divisor is not contained by the dividend, place a cipher in the quotient, and 
annex two ciphers to the divisor, and bring down the next period for a dividend, and use 
the divisor, as thus increased, for finding the next quotient figure. 

3. If there is a remainder after finding the integer of the root, annex a period of three 
ciphers thereto, and proceed for the decimal of the root as if seeking for the integer, an- 
nexing a period of three ciphers to each remainder until the decimal is carried to as many 
places of figures as desired. 

Example. — Required the cube root of 47421875.6324. 

4742i875.632400 ( 361. 959+. 
27 Ans. 

3 2 X 300 = 2700)20421 
6 

16200 
6 2 X 3 X 30 = 3240 

6 3 = 216=19656 



36 2 X 300 = 388800 ) 765875 
1 

388800 
l 2 X 36 X 30 = 1080 

l 3 = 1 = 389881 

36 1 2 X 300 = 39096300 ) 375994632 
9 

351866700 
9 2 X 361 X 30 = 877230 

9 3 = 729 = 352744659 

3619 2 X 300 = 3929148300 ) 23249973400 
5 



19645741500 
5 2 X 3619 X 30 = 2714250 

5 3 = 125 = 19648455875 

36195 2 X 300 = 393023407500 ) 3601517525000 
9 

3537210667500 
X 36195 X 30 = 87953850 

9 3 ^= 729 = 35372 9 8622079 

64218902921 



EVOLUTION. 145 

Example. - - Required the cube root of 32768. Also, of 8489664. 



32768(32. 
27 Ans. 

32x300 = 2700 ) 5768 
2 

5400 
2 2 X 3X30=360 

23= 8 = 5768 



8489664(204. 
8 Ans. 



2 2 X 300 = 120000 ) 489664 
4 

480000 
4 2 X 20 X 30 = 9600 

4 3 = 64=489664 



General Rule for extracting the roots of all powers, or for finding 
any proposed root of a given number. 

1. Point off the given number into periods of as many figures 
each, counting from right to left, as correspond with the denomina- 
tion of the root required ; that is, if the cube root be required, into 
periods of three figures, if the fourth root, into periods of four fig- 
ures, &c. 

2. Find the first figure of the root by inspection or trial, and place 
it at the right of the number, in the form of a quotient ; raise this 
quotient figure to a power corresponding with the denomination of 
the root sought, and subtract that power from the left hand period, 
and to the remainder bring down the first figure of the next period, 
for a dividend. 

3. Raise the root thus far found (the quotient figure) to a power 
next inferior in denomination to that of the root required, multiply 
this power by the number or index figure of the root required, and 
take the product for a divisor ; find the number of times the divisor 
is contained in the dividend, and place the result (except that the 
remainder is large, diminished by one or two units) in the quotient, 
for the second figure of the root. 

4. Raise the root thus far found (now consisting of two figures) to 
a power corresponding in denomination with the root required, and 
subtract that power from the two left hand periods, and to the re- 
mainder bring down the first figure of the third period, for a divi- 
dend ; find a new divisor, as before, and so proceed until the whole 
root is extracted. 

Example. — Required the fifth root of 45435424. 

45435424(34. Ans. 
3 5 = 243. 

3*X5)2113 

34 5 = 45435424 

13 



146 ARITHMETICAL PROGRESSION. 

Example. — Required the fifth root of 432040. 0o54. 

432040.03540 ( 13.4 -f. Ans. 

1 5 = 1 

l 4 X 5 ) 33 
13 5 = 371293 



13 4 X5) 607470 
13. 4 5 =43204003424 

116 

For instructions touching special cases, see Notes relative to the 
extraction of the square root, and to the extraction of the cube root. 

The As/ of the a/ of any number = ts/ of that number 
" Vofthe^ = /y. 
" a/ of the a/ of the a/ = £/. 

" % of the Ay = Ay. 

" a/ of the aV = a7, &c. 



AEITHMETICAL PROGRESSION. 

A series of three or more numbers, increasing or decreasing by 
equal differences, is called an arithmetical progression. If the num- 
bers progressively increase, the series is called an ascending arith- 
metical progression; and if they progressively decrease, the series is 
called a descending arithmetical progression. 

The numbers forming the series are called the terms of the 
progression, of which the first and the last are called the extremes, 
and the others the means. 

The difference between the consecutive terms, or that quantity by 
which the numbers respectively increase upon each other, or decrease 
from each other, is called the common difference. 

Thus, 3, 5, 7, 9, 11, &c, is an ascending arithmetical progression, 
and 11, 9, 7, 5, 3, is a descending arithmetical progression. In these 
progressions, in both instances, 11 and 3 are the extremes, of which 
11 is the greater extreme, and 3 is the less extreme. The numbers 
between these, (9, 7, 5,) are the means. 

In every arithmetical progression, the sum of the extremes is 
equal to the sum of any two means that are equally distant from the 
extremes ; and is, therefore, equal to twice the middle term, when 
the series consists of an odd number of terms. Thus, in the fore- 
going series, 3-f-ll=5-{-9 = 7X2. 

The greater extreme, the less extreme, the number of terms, the 



ARITHMETICAL PROGRESSION. 147 

common difference, and the sum of the terms, are called the Jive prop- 
erties of an arithmetical progression, of which, any three being given, 
the other two may be found. 

Let s represent the sum of the terms. 
" E " the greater extreme. 
" e " the less extreme. 
11 d " the common difference. 
" n " the number of terms. 

The extremes of an arithmetical progression and the number of terms 
being given, to find the sum of the terms. 

(E + e) X n 

2 == sum °f the terms. 

Example. — What is the sum of all the even numbers from 2 to 
100, inclusive ? 

102 X 50 -r- 2 = 2550. Ans. 

Example. — How many times does the hammer of a common clock 
strike in 12 hours % 

(1 + 12) X 12 -r- 2 = 78 times. Ans. 
( — - — + 1 j x — 9 — — sum °f * ne terms. 



(BX2-k-1X4X^ = sum of the terms. 



(2e-f-n — iXd) Xi« = sum of the terms. 

The greater extreme, the common difference, and the number of terms 
of an arithmetical progression being given, to find the less extreme. 



E — (d X n — 1) = less extreme. 

Example. — A man travelled 18 days, and every day 3 miles far- 
ther than on the preceding ; on the last day he travelled 56 miles ; 
how many miles did he travel the first day ? 



56 — (18 — 1 X 3) = 5 miles. Ans. 

— ( ) = less extreme. 

n \ 2 / 

s 

— X 2 — E = less extreme. 



148 ARITHMETICAL PROGRESSION. 

V (E X 2 -\- dy- — s X d X 8 -j- ^ = less extreme, when 
2 
<\/ (2 E -j- rf ) 2 — 8 s </ is equal to, or greater than d. 



/s/ (2 E -|- d) 2 — 8 s d ^ d= less extreme, when 
_ 

As/ (2 E -j- d)" — 8 s d is less than e?. 



/%/(2 e >W) 2 -f- 8 s e? — d= greater extreme 



dXn — 1 -f- e = greater extreme. 



s n — 1 X d 

— -j- 5 = greater extreme. 

2 s —- n — e = greater extreme. 

The extremes of an arithmetical progression and the common difference 
being given, to find the number of terms. 

E — e -=- c? — |— 1 = number of terms. 

Example. — As a heavy body, falling freely through space, de- 
scends 16^ feet in the first second of its descent, 48 T 3 2- feet in the 
next second, 80 T 5 2 in the third second, and so on ; how many sec- 
onds had that body been falling, that descended 305 T 7 2- feet in the 
last second of its descent ? 

305 T 7 F — 16 T V = 289£ -T- 32£ = 9 + 1 = 10 seconds. Ans. 

V (2e^d) 2 -{-8 s d — d — e -j- e? -[- 1 = number of terms. 
2 



2 5 -j- E -f- as/ (2 E -j- d)2 — 8 sd-\-d= number of terms when 
2 
/\/ (2 E -|- d) 2 — 8 s d is equal to, or greater than d. 



2s-7-E-(- \/(2E-|-d) 2 — 8 sd,W = number of terms when 
_ 

V(2E-4-d) 2 — 8sd is less than d. 
«X2 



E+, 



= number of terms. 



ARITHMETICAL PROGRESSION. 149 

The extremes of an arithmetical progression, and the number of terms 
being given, to find the common difference. 

E — e 

7 = common difference. 

n — 1 

Example. — One of the extremes of an arithmetical progression is 
28 and the other is 100, and there are 19 terms in the series ; re- 
quired the common difference. 

100 ^ 28 -r- 1^19 = 4. Ans. 

AX2 \ 
E — e-f-l s I — 1 ) = common difference. 

2s^-n — 2e 

-^ = common difference. 

n — 1 

2E— (2s + n) 

i = common difference. 

n — 1 

Example. — The less extreme of an arithmetical progression is 28, 
the sum of the terms 1216, and the number of terms 19 ; required 
the 7th term in the series, descending. 

1216 X 2 -H 19 = 128 = sum of the extremes. 
128 — 28 = 100 = greater extreme. 
100 — 28 = 72 = difference of extremes. 



72 -r- n — 1 (18) = 4 = common difference. 



100 — (7 — 1 X 4) = 76 = 7th term descending. Ans. 
Required the 5th term from the less extreme, in an arithmetical 
progression, whose greatest extreme is 100, common difference 4, 
and number of terms 19. 



100 — (19 — 5 X 4) = 44. Ans. 

To find any assigned number of arithmetical means, between two given 
numbers or extremes. 

Rule. — Subtract the less extreme from the greater, divide the 
remainder by 1 more than the number of means required, and the 
quotient will be the common difference between the extremes ; 
which, added to the less extreme, gives the least mean, and, added 
to that, gives the next greater, and so on. 



Or, E — e-r-m-\-l = d, E being the greater extreme, e the less 
extreme, m the number of means required, and d the common differ- 
ence. 

Ande-^d, e + 2 d, e-\- 3 d, &c. ; or, E — d, E — 2 d, E — 3 d, 
&p., will give the means required. 

13* 



150 GEOMETRICAL PROGRESSION. 

Example. — Required to find 5 arithmetical means between the 
numbers 18 and 3. 

18 — 3 = 15 -r- 6 =2^, and 
3 + 2i=.54 + 2J = 8 + 2i = 10a + 2i = 13 + 2i=-154. 
5£, 8, 10£, 13, 15£, therefore, are 5 arithmetical means, between the 
extremes, 3 and 18. 

Note. — The arithmetical mean between any two numbers may be found by dividing 
the sum of those numbers by 2 ; thus, the arithmetical mean of 9 and 8 is (9 -f- 8) -j- 2 =.8£. 



GEOMETRICAL PROGRESSION. 

«• 

A series of three or more numbers, increasing by a common mul- 
tiplier, or decreasing by a common divisor, is called a geometrical 
progression. If the greater numbers of the progression are to the 
right, the progression is called an ascending geometrical progression, 
but, on the contrary, if they are to the left, it is called a descending 
geometrical progression. The number by which the progression is 
formed, that is, the common multiplier, or divisor, is called the 
ratio. 

The numbers forming the series are called the terms of the pro- 
gression, of which the first and the last are called the extremes, and 
the others the means. The greater of the extremes is called the 
greater extreme, and the less the less extreme. 

Thus, 3, 6, 12, 24, 48, is an ascending geometrical progression, 
because 48 is as many times greater than 24, as 24 is greater 
than 12, &c. ; and 250, 50, 10, 2, is a descending geometrical pro- 
gression, because 2 is as many times less than 10, as 10 is less than 
50, &c. - ' 

In the first mentioned series, (3, 6, 12, 24, 48,) 48 is the greater 
extreme, and 3 is the less extreme; the numbers 6, 12, 24 are the 
means in that progression. 

So, too, of the progression 250, 50, 10, 2 ; 250 and 2 are the ex- 
tremes, and 50 and 10 are the means. 

In the first mentioned progression, 2 is the ratio, and in the last, 
or in the progression 2, 10, 50, 250, 5 is the ratio. 

In a geometrical progression, the product of the two extremes is 
equal to the product of any two means that are equally distant from 
the extremes, and, also, equal to the square of the middle term, 
when the progression consists of an odd number of terms. 

Thus, in the progression 2, 6, 18, 54, 162 ; 162 X 2 = 54 X 6 
= 18 X 18. 

When a geometrical progression has but 3 terms, either of the 



GEOMETRICAL PROGRESSION. 151 

extremes is called a third proportional to the other two ; and the 
middle term, consequently, is a mean proportional between them. 

Thus, in the progression 48, 12, 3, 3 is a third proportional to 48 
and 12, because 48 divided by the ratio = 12, and 12 divided by the 
ratio =b 3 ; or 3 X ratio = 12, and 12 X ratio = 48 : 12 is the mean 
proportional, because 12 X 12 = 48 X 3. 

Of the 5 properties of a geometrical progression, viz., the greater 
extreme, the less extreme, the number of terms, the ratio, and the sum 
of the terms, any three being given , the other two may be found 

Let 5 represent the sum of the terms. 

E " the greater extreme. 

e " the less extreme. 

r " the ratio. 

n " the number of terms. 

n when affixed as an index or exponent, represent that the 
term, number, or quantity, to which it is affixed, is to be raised 
to a power equal to the number of terms in the respective progres- 
sion, &c. 

Any three of the five parts of a geometrical progression being given, to 
find the remaining two parts. 

E — e 

r- -4- E = sum of the terms. 

r — i ' 

— sum of the terms. 



r— 1 
r» X c — e 

" r— 1 

E— (E-f- r^T) 

E — e 



sum of the terms. 

|- E = sum of the terms. 
J- E = sum of the terms. 



Example. — The greater extreme of a geometrical progression is 
162, the less extreme is 2, and there are 5 terms in the progression ; 
required the sum of the series. 

^1^^ = 80 + 162 = 242. Ans. 



= greater extreme. 



152 GEOMETRICAL PROGRESSION. 

— X«= greater extreme. 



r n " 1 Xe = gr 


eater extreme. 


sX r " _1 X*" — 1 




r n I 




s — (s — E)X 


r = less extreme, 
less extreme. 


5 — e 


- -j-r n-1 =less extr 


™ ••= ratio. 

5 ht 

F 

n 7y~ = ratio. 





sXr — l 

-}- 1 = r a ; n, therefore, is equal to the number of times 



sXr — l 
that r must be multiplied into itself to equal -f- 1. 



'Xr-1 ^ 



s _( 5 _E)Xr 

Example. — A farmer proposed to a drover that he would sell 
him 12 sheep and allow him to select them from his flock, provided 
the drover would pay 1 cent for the first selected, 3 cents for the 
second, 9 cents for the third, and so on ; what sum of money would 
12 sheep amount to, at that rate ? 

r n X e — e 

i — = 5, then 

312 x 1 — 1 

3TTT - = $2657.20. Arts. 

Note. — Ratio , cubed = ratio ; ratio 6 , squared = ratio , &c. 

When it is required to find a high power of a ratio, it is conven- 
ient to proceed as follows, viz. : write down a few of the lower or 
leading powers of the ratio, successively as they arise, in a line, one 
after another, and place their respective indices over them ; then 



GEOMETRICAL PROGRESSION. 



153 



will the product of such of those powers as stand under such indices 
whose sum is equal to the index of the required power, equal the 
power required. 

Example. — Required the 11th power of 3. 

12 3 4 5 

3 9 27 81 243 

Here 5 — |- 4 — {— 2 = 11, consequently, 

243 X 81 X 9 = 11th power of 3, or 

5 X 2 -J- 1 = 11, consequently, 
243 X 243 X 3 = 11th power of 3, or 

4 X 2 -f- 3 = 11, consequently, 
81 X 81 X 27 = 11th power of 3, or 

3 X 3 -\- 2 = 11, consequently, 

27 3 X 9 = r 11 = 177147. Ans. 

To find any assigned number of geometrical means, between two given 
numbers or extremes. 

Rule. — Divide the greater given number by the less, and from 
the quotient extract that root whose index is 1 more than the 
number of means required ; that is, if 1 mean be required, extract 
the square root ; if two, the cube root, &c, and the root will be the 
common ratio of all the terms ; which, multiplied by the less given 
extreme, will give the least mean ; and that, multiplied by the said 
root, will give the next greater mean, and so on, for all the means 
required. Or the greater extreme may be divided by the common 
ratio, for the greatest mean ; that by the same ratio, for the next 
less, and so on. 

Example. — Required to find 5 geometrical means between the 
numbers 3 and 2187. 

2187 -f- 3 = 729, and 4/729 = 3, then — 
3X3 = 9X3 = 27X3=81X3 = 243X3 = 729, that is, the 
numbers 9, 27, 81, 243, 729 are the 5 geometrical means between 
3 and 2187. 

Note. — The geometrical mean between any two given numbers is equal to the square 
root of the product of those numbers. Thus the geometrical mean between 5 and 20, = 
V (5X20) = 10. 



154 ANNUITIES. 



ANNUITIES. 

An annuity, strictly speaking and practically, is a certain sum 
of money by the year ; payable, usually, either in a single pay- 
ment yearly, or in half, half-yearly, quarter, quarter-yearly, &c, 
and for a succession of years, greater or less, or forever. Pensions, 
awards, bequests, and the like, that are made payable in fixed 
sums for a succession of payments, are commonly rated by the 
year, and denominated annuities. 

A current annuity that has already commenced, or that is to 
commence after an interval of time not greater than that between 
the stipulated payments, is said to be in possession. 

One that is to commence or cease on the occurrence of an 
indeterminate event, as upon the death of an individual, is a re- 
versionary, contingent, or life annuity. 

One that is to commence at a given period, and to continue for a 
given number of years or payments, is a certain annuity. 

One that is to continue from a given time, forever, is a perpetual 
annuity, or a perpetuity. 

Annuity payments do not exist fractionally : they mature, and 
exist only in that state, and are then due. 

A current annuity commences with a payment, and terminates 
with a payment. 

One current in the past is measured from a present included pay- 
ment, closes with an included payment, and is said to be in arrears 
or forborne, from a supposed cancelled payment one regular inter- 
val or time beyond. 

One current in the future is measured from the present to the 
first included payment of the series, and from thence is said to con- 
tinue to the close ; but if the interval from the present to the first 
included payment is equal to that between the successive pay- 
ments, it is supposed to continue from the present. 

Annuities in negotiation are adjusted, with regard to time, by 
interest, or discount, or both. 

The tables applicable to compound interest and compound dis- 
count are applicable in adjusting annuities at compound rates. 

To find the Amount of a Current Annuity in Arrears. 

Lemma. — The amount of an annuity that has been forborne 
for a given time is equal to the sum of the several payments that 
have become due in that time, plus the interest on each, from the 
time it became due, until the close of the time. 



ANNUITIES. 155 

Then the amount of an annuity of $100, payable in a single 
payment annually, but delayed of payment 4 years, allowing sim- 
ple interest at 6 per cent, on the payments, is 

100 X 1.18= 118 
100 X 1.12= 112 
100 X 1.06= 106 

100 X 1 = 100 = $436. 

And at 6 per cent, compound interest on the payments, it is 

100 X (1.06) 3 = 119.10 
100 X (1.06) 2 = 112.36 
100 X (1.06) 1 = 106.00 
100 X 1 =100.00 = $437.46. 

At 6 per cent, simple interest, when payable in half, half- 
yearly, it is 

50 X 1.21=60.50 
50 X 1.18 = 59.00 
50 X 1.15 = 57.50 
50 X 1.12 = 56.00 
50 X 1.09 = 54.50 
50 X 1.06 = 53.00 
50 X 1.03 = 51.50 
50 X 1 — 50.00 = $442. 

And at 6 per cent, compound interest per annum, when payable 
in half-yearly instalments, it is 

50 X (1.06) 3 X 1.03=61.34 

50 X (1.06) 3 =59.55 

50 X (1.06)2 x 1.03=57.86 

50 X (1.06)2 =56.18 
50 X (1.06)1 x 1.03 = 54.59 

50 X 1-06 =53.00 

50 X 1-03 =51.50 

50 X I- =50.00 = $444.02. 

From the foregoing, we derive the following general Rules : — 

Let P = annuity or yearly sum, 

r = rate of interest per annum, 

a = rate of discount per annum, 

n or n = nominal time of the annuity in full years, 

A = amount for the full years, 

D =present worth for the full years. 



156 ANNUITIES. 

When the annuity is payable in a single payment yearly, 
A = Pn (l + ^f^)t Simple Interest. 
A = P — ~. i Compound Interest. 

When payable in equal half-yearly instalments, 

A = P«(l-f- r -&f±± -f -f ), Simple Interest. 

A= P X ■ 1+r i. a " 1 x(l + j )' Compound Interest. 
When payable in equal third-yearly instalments, 

A = Pn ( 1 + r( "~ 1) + -f ) i Simple Interest. 

A=P (1+ ^ ) ~ 1 (l-f |), Compound Interest. 
When payable in quarter-yearly instalments, 

A = Pn ( 1 + r ~^Y 1)j r t )' Sim P le Interest - 
A — P (1+r) °~ 1 ( l -J- ip ), Compound Interest. 

When there are odd payments, to find the amount, S. 

When 1 half-yearly, S = A(l -f- i r) 

1 third-yearly, S = A(l + i r ) 

2 " S = A(l + fr) 



IP 

2 x. 

3 r> 



|P(i + H+iP 



P(8 + r)-^16. 
P(3 + IO-^ 4 - 



■ r3 -V JL ~3 / \^ 3 \ I 3 J I 

-A(l+fr) + P(6+r)^-9. 

1 quarter-yearly, S = A(l -f- £ r) -(- |P. 

2 " S = A(l + £r) 

3 . " S = A(l + fr) 

For any number of equal and regular payments at compound 

interest per interval between the payments, S = P' ( — — j, and 

for any number of equal and regular payments at simple interest 
per interval between the payments, S = P'n' (l -f- r{n ~ ) ; P' 
being a payment, n' or n ' the number of payments, and r' the rate 
of interest per interval between the payments. But this must not 
be confounded with compound interest annually, on payments oc- 
curring semi-annually, quarterly, &c. 

Example. — What is the amount of an annuity of $150, paya- 
ble in half, half-yearly, but delayed of payment 2 years and 72 
days, allowing compound interest per annum at 7 per cent. V 

150 x ( ~^ = $310.50, the amount for 2 years, if payable 
» in yearly payments, and 



ANNUITIES. 157 

-310.50X O- 1 ^) =$315.93, the amount for 2 years, if payable 
in half-yearly payments, and 

315.93 X C 1 -^^) =$320.29, the amount for 2 years and 72 
days, if payable in half-yearly payments. Ans. 

Example. — What is the amount of an allowance, pension, or 
award, of $100 a year, payable quarterly, but forborne 3| years, 
interest compound per annum at 6 per cent. ? 

H)0 X (1 '° 6 ) 6 ~ 1 X ( 1 -f 1 ^) = $325.52, the amount for 3 years, 

and 

325.52 (1+ .03) 4- 100X8.06 ~ 16= $385.66. Ans. 

Example. — What is the amount of $100 a year, payable in 
quarterly payments, and in arrears 4 years, interest being com- 
pound per quarter-year, at 6 per cent, a year ? 

25 [(1 + x) 16 — l] X -^ • B y tabular powers of (1 -f- r), page 

'125, =$448.30. Ans. 

To find the Present Worth of an Annuity Current. 

Lemma. — The present worth of an annuity that is to continue 
for a given time is equal to that sum of money, which, if put at 
interest from the present time to the close of the payments, will 
amount to the amount of the payments at that time ; and therefore, 
the times being full, is equal to the sum of the several payments, 
discounted, respectively, at the rate of interest for their respective 
times. 

Note. — If the foregoing proposition is tenable, it follows, since simple 
interest is due and payable annually, that the true present worth of an annuity 
having more than one year to run cannot be found by simple interest and dis- 
count. By simple interest and discount, at 6 per cent., predicating the rule 
upon the foregoing lemma, the amount of $100, payable annually, and in 
arrears for 4 years, is $436; and the present worth, at 6 per cent., is 

100 100 100 , 100 _ 0/m 

But $349 at 6 per cent, interest for 4 years, with the payments of interest annu- 
ally, will amount to $440.30 ; and at interest simply for 4 years it will amount 
to only $432.76. 

Then the present worth of an annuity of $100, payable in a 
sintxle payment yearly, and to continue 4 years, or to become due 
1, 2, 3, and 4 years hence, interest and discount being compound 
per annum, and each at 6 per cent. -=. 
14 



158 



ANNUITIES. 



P P P 



(1+r) 4 . (1+r) . 

100 X (1.06) 8 = 119.10 
100 X (1.06) 2 = 112.36 
100 X (1.06) —106.00 
100 X 1 =100.00 — 43 7.46 — (1.06)* = $346.51. 

And interest at 6 per cent, and discount at 10, both compound, it is 
100 X (1.06) 8 = 119.10 ■ 

100 X (1.06) a = 112.86 
100 X 1-06 =106.00 
100 X 1 =100.00= 437.46 -f- (1.10) 4 = $298.79. 

Therefore, when the annuity is payable in a single payment 
yearly from the present time, 

D = P ~ ,— — — i when r and a are equal. 

When payable in half-yearly payments, 

D='Px ff±f ^=rX(l + iO- 

ril + a) i*s 

When payable in third-yearly payments, 

■p_ PxC(l + r) lt — l]x(l+jr^ 
ra + af 
When payable in quarter-yearly payments, 
p __ P [(l+r) n -l](l+fr) . 
r(l + of 
When there are odd payments, to find the present worth, S. 
There being a half-yearly, S= 1 ^- a -f f ^_. 
1 third-yearly, 8=^ + ^. 

o li g— . _P_.i_ 2P(l+8r) . 

l+§a I 8(l + fa) 

1 quarter-yearly, S= r ^+ r g i - 

«» « s — d i PH+lr). 

* °— l + ia~T~ 2(1-H«) 

3 " 



q B i fP(l+lr). 

b -ffTa + 



l+f« 



For any number of equal payments, at equal intervals between 



the payments, S = P' X 



_ P' ss a + r'f-l . 



ra + a') 



P' being a payment, n ' the 



ANNUITIES. 159 

number of payments, and r' and a' the rates per interval between 
the payments. 

Note. — Since ( -~t^ — ^— is the co-efficient of P, for its present worth, at 

compound interest and discount, for the time » , at the rates r, a, it follows 
that tables of co-efficients of P for its present worth, at given rates, for any 
number of years, may be easily made. Thus (1.06 4 — l)-f-l.06 4 x .06 = 3.46511, 
the co-efficient of an annuity, P, for 4 years' continuance, interest and discount 
being compound per annum, at 6 per cent.; and (1.06 2 — l)-j-(1.06 2 X-06)=: 
1.83339, the co-efficient for 2 years, &c. 

If the annuity is deferred, then the difference of two of these co-efficients 
(one of them that for the time deferred, and the other that for the sum of the 
ti.ne deferred and the time of the annuity) will be the co-efficient of P for 
its present worth. Thus 3.46511 — 1.83339= 1.63172, the co-efficient of an annuity, 
P, for its present worth, when it is to commence two years hence, and to con- 
tinue 2 years, interest and discount being compound per annum, at 6 per 
cent, each ; orD= 1.63172 P. 

In like manner, tables of other co-efficients, such as the formulae suggest, 
may be made that will greatly assist in calculating annuities. 

Example. — What is the present worth of an award of $500 a 
year, payable in half-yearly instalments, the 1st payment to mature 
6 months hence, and the annuity to continue three years ; interest 
and discount being 7 per cent, compounded yearly ? 

500 X [(1.07) 3 — 1] X (l.f) 

07xil0ir ^-^=mS5.1S. Ans. 

Example. — What is the present worth of an annuity of $100, 
payable in half-yearly payments, and to continue 1^ years; interest 
and discount being 6 per cent, per annum ? 

100 X [1-06 — 1] X l-~ 



.06 X 1.06 



95.755, and 



95.755 , 50 , 

ToF + ro3 = $141 - 51 - Ans ' 

Example.— -What is the present worth of an annuity of $500, 
payable in semi-annual instalments, and to continue 10i years, 
interest and discount being compound per annum, the former at 6 
per cent., and the latter at 8 ? 

500[(l,06) 1Q -l](l4 6 ) 50Q 



.06(1.08)" (l.f) ~ r 2(l.f) 

A 250 

1.08 10 X 1-04 + 1.04 ~ AnS ' 



160 ANNUITIES. 

By tabular powers of 1 -(- r, page 125 : — 
- •' / ' - = 83052.64, the present worth for 10 years' con- 

• Ub X ^•i.Oov'A 

tinuance, if payable in yearly payments, and 

3052.64 X 1-015 — $3098.43, 

the present worth for 10 years' continuance, if payable in half- 
yearly payments, and 

3098.43 -f- 1.04 -|- 500-1-2 X 1.04 =$3219.64. Am. 

When the interval of time from the present to the 1st payment 
is shorter than that between the consecutive payments, and the 
annuity is payable in a single payment yearly, 

• A= P E q-K-i3(i + a 

r 

dr' 



D = 



A _ P[(l+Q P -l](l+ 3 -f) 

(l+ay-^l+^gp*) " (l+a)^(l + ^»)' 
d being the time in days from the present to the 1st payment. 

So, if the annuity is payable in half-yearly, third-yearly, or quar- 
ter-yearly instalments, multiply by 1 -j- \ r, 1 -j- £r, or 1 -j- -fr, as 
before directed ; and if there are odd payments proceed for the 
present worth, S, as already directed. 

Example. — Required the present worth of an annuity of $100, 
payable yearly, to commence 4 months hence, and to continue 4 
years ; interest and discount being 6 per cent, annually. 

100X(1.06 4 -l)x(l. 4 -^f) 



.06 X 1-06 3 X (l.- 06 ^- 4 ^ 



=i $860,24. Am. 



To find the Present Worth of a Deferred Current Annuity, or of 
an Annuity in Reversion. 

When the annuity is payable in a single payment yearly, and the 
deferred time embraces full years only, 

(I _L r \ n i 

D = P — — r , , ,» , n' being the deferred time. 

If it is payable in half-yearly, third-yearly, or quarter-yearly 
instalments, multiply by 1 + £ r, 1 -|- £ r, or 1 + § r, as already 



ANNUITIES. 161 

directed ; and, if there are odd payments, find the present worth, S, 
as already directed. 

Example. — What is the present worth of an annuity of $150, 
payable yearly, to commence 2 years hence, and to continue 4 
years ; interest and discount being compound per annum, at 6 per 
cent. ? 

150 X (1-06 4 — 1) -r- -06 X 1.06 6 =: $462.59. Ann. 

Example. — Required the present worth of an annuity of $500, 
payable in semi-annual instalments, to commence 2^- years hence, 
and to continue 6 years ; allowing compound interest and discount 
annually at 7 per cent. 

07 



500 X (1.07 6 — 1) X I-— 



= $2046.44. Ans. 
.07 X 1-07 3 X l-'y 



.07 



Example. — Required the present worth of an allowance, 
pension, or award of $125 a year, payable in half every half-year, 
to commence 7 months 24 days hence, and to continue 6£ years ; 
interest and discount being compound per annum at 5 per cent. 

125 X (1.05° -1)X 1-0125 125 

.05 X 1-05 6 X 1-03247 X 1-025 T 2X 1-025 

Or ^-^ ^fPlZ 1 ^ = $634.47, the present worth for 6 
.05 X (l-0o) 6 l 

years' continuance, if payable in yearly instalments ; and 
634.47 X 1.-^- = $642.40, 

the present worth for 6 years' continuance, if payable in half-yearly 
instalments ; and 

642.40 -f- (l + -^~) = 622.20, 

the present worth for 6 years' continuance, if payable in half-yearly 
instalments, and to commence 7 months, 24 days hence ; and 

622.20 4-(i+|) +2 - 3 ^ = $668. Ans. 



To find the Present Worth of a Perpetuity. 

Lemma. — The present worth of an annuity to commence one 
year hence, and to continue forever, is expressed by that sum of 
money whose interest for 1 year is equal to the amount of the 
14* 



162 ANNUITIES. 

annuity for 1 year ; and so, pro rata, for perpetuities otherwise 
regularly affected. 

Then when the annuity is to commence 1 year hence, and is 
payable in a single payment yearly . . . D == P —- r. 

Payable in half-yearly instalments . . D .= — 



Payable in third-yearly instalments . . D = 
Payable in quarter-yearly instalments . D =: 



r 

r 

PCl+jr) 
r 



Example. — What is the present worth of a perpetuity of $150 
a year, payable in a single payment yearly from the present time ; 
interest at 6 per cent ? 

150 -f- .06 = $2500. Ans. 

Example. — What is the present worth of a perpetuity of $150 
a year, payable in semi-annual instalments, and to commence 4 
months hence ; interest 7 per cent ? 

PO+jr) 07(12-4) =$218786 . Am _ 

r ' 12 

Example. — Required the present worth of a perpetuity of 
$400 a year, payable in quarterly payments, and to commence 6 
years hence ; interest and discount being 5 per cent., compound 
per year. 

P(l+fr) 400 Xl 3 -^- 

T> - - = ^- == $6081.65. Ans. 

D — r (l + ay .05X1-05 6 



The Amount, Time, and Rate given, to find the Annuity. 
When payable in a single payment yearly from the present time, 

* = (l+7f=T i *****> P = [-(l+r)- -!](!+*■) » 
third-yearly, P = (1+ ^ £ + r)n _ fj i quarterly, 



(l + l'-)[(l + '-)*-l] ' 



ANNUITIES. 163 

and so, pro rata, for other fractional units of the integral unit. 
Therefore (i + r )« — 1 = ^, or — , r — 

Ar 

or f(+M' k 

Example. — What annuity, payable in quarterly payments 
from the present time, will amount to $3000 in 12 years; interest, 
being compound per annum, at 8 per cent. ? 

3000 X -08 -^- [(108 12 — 1) X 1. 3 -^-H] = $153.48. Ans. 

Example. — What length of time must a current annuity of 
$400, payable in quarterly payments, remain unpaid, that it may 
amount to $2500 ; interest being 7 per cent, yearly ? 

2500 V 07 
. ' „ 7 = .4263094 = 5-4- years, and 5 years by table of 

400 Xl.°-j5- 7 

£+*-!- -40 2552 , therefore (^_ t) £_ 308 
days, 5 years, 308 days. ^ins. 



TAe Present Worth, Time, and Rate given, to find the Annuity. 
When payable in a single payment yearly from the present time, 

p - (i+r)t- 1 » hrif-y«iy. p - [(1+r) .LiV+ir) ; third " 

yearly, P = [(1 +r ^ + f + , r) J quarter-yearly, P = 

Dr(l 4- r) n P 

[(l+ry-rKl + fr) ' &c - Therefore, (1 -fr)>;= p -— ^ = 

F(l+jfr) _ P(l + |r) 
• P(l_[_^.)_Dr~ P(l+fr) — Dr ' ' 

Example. — What annuity, payable in half-yearly instalments, 
and to continue 3 years, is at present worth $1335.13 ; discount and 
interest being compound per year, at 7 per cent ? 

1335.13X-07X1.07 S 



07 



(1.07 3 — l)Xl^f 



= $500. Ans. 



164 ANNUITIES. 



OF INSTALMENTS GENERALLY. 

Any certain sum of money to be paid on a debt periodically 
until the debt is paid is called an instalment ; and a debt so made 
payable is said to be payable by instalments. 

Let D = principal or debt to be paid, 

n = number of years in which the debt is to be paid, 

r = rate of interest per annum, 

p = instalment or periodical payment. 

When the instalments are payable yearly, and the debt is at 
interest, 

Dr(l+r) - _ p ^ [(l + r)°-l] 

^-(l_|_ r )n_l iV-T r ) — p — Dr ' -"— r (l+r) n ' 

When payable half-yearly, 

Dr(l -f r)° 

^-2[(l+rr-l](l + ir] ; 

n , rV ._ K l + fr)+P . D _ 2pr(l+r)°-l](l+jr) 
S 1 "T r '-[ 1 ,(l+ir)+.j,]_I>r' lU - — r(I+r)» 

When the debt is not on interest, and the instalments are pay- 
able yearly, 

^— (i_L. r )n__i ' Cl+^) — p ' U — r 

Example. — What yearly instalment will pay a debt of $4000 
in 4 years, the debt being on interest the while, at 6 per cent, 
annually ? 

4000 X .06 X 1-06 4 -f- (1.06 4 — 1) =$1154.37. Ans. 

Example. — What semi-annual instalment will pay a debt of 
$4500 in 3 years, the debt bearing interest at 7 per cent, yearly ? 

4500—07 X 1.07* = $842.62. Ans. 



2 X (1-07 3 — 1) X 1.0175 

When the debt is on interest, and is payable in equal yearly in- 
stalments, p = D(l -f" rn) -f- w(l -f- ^f 11 ), at simple interest ; but 
simple interest is not strictly applicable to instalments. Sec 
Note, p. 157. 



ANNUITIES. 165 

When a debt has been diminished at regular intervals by the 
payment of a constant sum, to find the remaining debt at the close 
of the last payment. 

When the debt is on interest, and the payments have been made 
yearly from the date of the debt, 

d - P— (P— Dy, )Q +0° . n ■ r ^_ P — dr 
r ,^-f-r) - p _ J)r 

__ P r (l -f r )° — dr 
P~ (l+r)--l ' 

When the payments have been made half-yearly, 

d=p+p(l+ir) — O + O-C^ + pCl+ir) — Dr]-L. r ,&c 

Example. — On a debt of $1000, drawing interest the while 
at 8 per cent, a year, there has been paid yearly, from the date of 
the debt, $200 for 6 years : required the unpaid debt at the close 
of the last payment. 

[200 — 1.08 6 (200 — .08 X 1000)] -f- .08 = $119.69. Ans. 

Example. — On a note of hand for $1000, and interest from 
date, at 8 per cent, annually, the following payments have been 
made; viz., $100 at the close of every half-year from the date of 
the note, for 6 years. How much remained' unpaid at the close 
of the last payment ? 

[204 — 1.08 6 (204 — .08 X 1000)] -f- .08 = $90.34. Ans. 

Note. — In the foregoing, I have treated the terms annual interest and inter- 
est payable annually as synonymous in meaning with the terms compound in- 
terest and compound interest per annum, and they are so in equity and in fact; 
besides, simple interest is inapplicable, in equity, to instalment payments. If 
the debtor stipulates to pay the interest annually on a debt, and abides his con- 
tract, he will pay it when it becomes due, and it then becomes a principal in 
the hands of the creditor, to be let, it is fair to suppose, upon as favorable 
terms to himself as he let the principal which grew it: whereby he realizes 
equal to compound interest per annum on the first principal : moreover, if 
the debtor withholds it from the creditor, it is fair to suppose that he considers 
it of as much worth to himsolf as a like part of the principal. 



166 PERMUTATION. 



PERMUTATION. 

Permutation, in the mathematics, has reference to the greatest 
number of unlike relative positions, that a given number of things, 
either wholly unlike, or unlike only in part, may be placed in. It 
considers the number of changes, therefore, that may be made, in 
the arrangement of the things, under different given circumstances. 

To find the number of changes that can be made in the order of arrange- 
ment of a given number of things, when the things are all different. 

Rule. — Find the product of the natural series of numbers, from 
1 up to the given number of things, inclusive ; and that product will 
be the number of changes or permutations that may be made. 

Example. — In how many different relative positions may 12 per- 
sons be seated at a table ? 

1X2X3X4X5X6X7X8X9X10X11X12 = 

479,001,600. Ans. 

To find the number of changes that can be made in the order of arrange- 
ment of a given number of things, when that number is composed of 
several different things, and of several which are alike. 

Rule. — Find the number of changes that could be made if the 
things were all unlike, as in first example. Then find the number 
of changes that could be made with the several things of each kind, 
if they were unlike. Lastly, divide the number first found by the 
product of the numbers last found, and the quotient will be the 
number of permutations or changes that the collection admits of. 

Example. — Required the number of permutations that can be 
made with the letters a, bb, ccc, dddd, = 10 letters. 

1X2X3X4X5X6X7X8X9X10 = 3628800 

1 X 2 X 6 X 24 = 288 ~~ 12 > bUU - Ans - 

To find the number of permutations that can be made with a given num- 
ber of different things, by taking an assigned number of them at 
a time. 

Rule. — Take a series of numbers beginning with the number of 
things given, and decreasing by 1 continually, until the number of 
terms is equal to the number of things that are to be taken at a 
time ; then will the product of the series be the number of changes 
that may be made. 



COMBINATION. 167 

Example. — What number of changes can be made with the 
numbers 1, 2, 3, 4, 5, 6, taking three of them at a time 1 

6 — 1 = 5, 5 — 1 = 4, then 6X5X4 = 120. Ans. 
What number, by taking 4 of them at a time ? 
6X5x4X3 = 360. Ans. 

Example. — Arrange the three letters a, b, c, into the greatest 
number of permutations possible. 

abc, acb, bac, bca, cab, cba, = 6 permutations. Ans. 

Example. — Arrange the four letters a, b, a, b, into the greatest 
number of permutations possible. 

abab, aabb, abba, bbaa, baba, badb, = 6 permutations. Ans. 



COMBINATION. 

Combination, in the mathematics, has reference to the number of 
unlike groups, which may be formed from a given number of differ- 
ent things, by taking any assigned number of them, less than the 
whole at a time. It does not regard the relative positions of the 
things, one with another, in any of the collections or groups. But 
it exacts that each group, in all instances, shall have the assigned 
number of members in it, and that, in every group, in every instance, 
there shall be a like number of members. It exacts, therefore, that 
no two groups shall be composed of precisely the same members. 

To find the number of combinations that can be made from a given 
number of different things, by taking any given number of them at a 
time. 

Rule. — Take a series of numbers beginning with that which is 
equal to the number of things from which the combinations are to 
be made, and decreasing by 1, continually, until the number of 
terms is equal to the number of things that are to be taken at a time, 
and find the product of those numbers or terms. Then take the natural 
series, 1,2, 3,4, &c, up to the number of things that are to be taken 
at a time, and find the product of that series. Lastly, divide the 
product first found by the product last found, and the quotient will 
express the number of combinations that can be made. 

Example. — What number of combinations can be made from 8 
different things, by taking 4 of them at a time? 
8X7X6X 5 _ 1680 __ 
1X2X3X4" 24 — 70 - Ans ' 



56. Ans. 



168 COMBINATION. 

What number, by taking 5 of them at a time ? 

8X7X6X5X4 6720 

1X2X3X4X5 - 120 
What number, by taking 3 of them at a time ? 

8 X 7 X 6 _ 336 en 

TX2XI T^ 56 - Ans ' 

Example. — What number of combinations can be made from 5 
different things, by taking three of them at a time ? 

5X4X3 60 

1X2X3" 6" 10 - Ans ' 

What number, by taking 2 of them at a time ? 

±Xi-?>-10. Ans. 
1X2 2 

Example. — Form 5 letters, a, b, c, d, e, into 10 combinations of 2 
letters each ; that is, into 10 unlike groups of two letters each. 

ab, ac, ad, ae, be, bd, be, cd, ce, de. Ans. 

Form them into the greatest number of combinations possible, in 
collections of three each. 

abc, abi, abe, acd, ace, ade, bed, bee, bde, cde. Ans. 



PROBLEMS. 169 

PROBLEMS. 

Prob. I. — The sum and difference of two numbers given, to find the 
numbers. 

Let a = the greater number, 

b = the less number ; then — 

2 ^ 

a-\-b-\-a^rb 

- = a, and a — a ^rb = b. 



2 

Prob. II. — The sum of two numbers and their product given, to find 
the numbers. 

V (a -j- b) 2 — (a X b X 4) = a^b, and 

a-\-b — a^nb 

o == b, aQ d ° ~f~ a ^ ° = a ' 

Prob. III. — The difference of two numbers and their product given, 
to find the numbers. 

a/ {a^bf-\-(aXbX±) = a + b, and 

a-\-b — a^rb 

2 = b, and b -\- a ^r b = a. 

Prob. IV. — The sum of two numbers and the sum of their squares 
given, to find the numbers. 

*/ ( c pJ r b 2 )X2— (a-\-b) 2 — a^b, thence, by Prob. I. 

Prob. V. — The difference of two numbers and the sum of their 
squares given, to find the numbers. 

V(fl 2 -j-J 2 )X2-(a^A) 2 = 4S, thence, by Prob. I. 

Prob. VI. — The sum of two numbers and the difference of their 
squares given, to find the numbers. 

a 2 ^rb 2 a-\-b — a^b 

j— 7 =a^b; n = b; b-\-a^nb = a; or 

a-f-o A ' 

( fl +J) X 2 = 5 ' and a + b~b=a. 

Prob. VII. — The difference of two numbers and the difference of 
their squares given, to find the numbers. 

a 2 ^b 2 a-\-b — a^b. 

r=a-\-b; o = b; b -4- a >s> b ^= a. 

15 



170 



PROBLEMS. 



Prob. Vm. — The product of two numbers and the sum of their 
squares given, to find the numbers. 



A /(a 2 -+-b 2 — aXbX2) = a^b, and 

V(fl 2 -|-5 2 + flX^X2) = a + i, and 

a-\-b — a^b 

9 = b, and b-\-a^r b = a. 

Prob. IX. — The sum of two numbers, and the product of those 
numbers plus the square of one of the numbers, in another sum given , 
to find the numbers. 

aXb-\-& 

a + b = b; a-\-°-b = a 

Prob. X. — The product of two numbers and the relation of those 
numbers to each other given, to find the numbers. 

,( aXb\ 
\ rXr / ) Xr = a, or Xr'=b; r being the term in relation 

representing the greater number, and r 1 being the term in relation 
representing the less. 

Prob. XI. — The sum of the squares of two numbers, and the rela- 
tion of those numbers to each other given, to find the numbers. 



V(jjr^)xr = 



a, or X i* = b. 



Prob. XII. — The sum of three numbers which are in arithmetical 
progression, and the sum of their squares given, to find the numbers. 

Let a = the greatest number, 
b = the middle number, 
c = the least number ; then 

a-\-b-\-c 

q = b, and a-\-b~\~c — 5 = « -|- e. 

\Z(a 2 + c 2 )X2 — (a + c) 2 =:a^c. 

a-\-c — a ^ c 

o = c, and c -j- a on c = a. 

Example. — The sum of three numbers which are in arithmetical 
progression is 18, and the sum of their squares is 140 ; what are 
the numbers ? 



PROBLEMS. 171 

18 -r- 3 = 6 = b, and 18 — 6 = 12 = sum of a and c . 
140 — 6 2 = 104 = sum of a 2 and c 2 . 
V (104 X 2 — 12 2 ) = 8 = a — c. 
12 — 8 = 4 -f- 2 = 2 = c, and 2 -f- 8 = 10 = a; the numbers, 
therefore, are 2, 6, and 10. J.ns. 

Note. — Half the sum of the first and third of three numbers forming an arithmetical 
progression is equal to the second number. 

Prob. XIII. — The sum of three numbers which are in arithmetical 
progression added to the sum of the greatest and least, and the sum of 
the squares of the numbers given, to find the numbers. 

a-\-b + c-\-a-\-c 
? = b, and 

a-\-b-\-c-\-a-\-c — b 

o = « + c ; thence, by Prob. XII. 

Prob. XIV. — The sum of three numbers which are in arithmetical 
ft ogression added to the sum of twice the greatest and twice the least, 
and the sum of the squares of the numbers given, to find the numbers. 

Za-\-3c-\-b 
~ = b, and 

Za + b + Zc—b 

o = a -\- c; thence, by Prob. XII. 

To find the altitude of an equilateral four-sided pyramid, the slant 
height and side of the base being known. 

*J (S 2 — h A 2 ) = h ; S being the slant height, A a side of the base, 
and h the altitude. 

V(F 2 — (i A) 2 X 2) = h ; F being the linear edge. 

V (F 2 —f)=h;f being half the diagonal. 

V(F-AX .7071 2 )* = A; 

To find the altitude of the frustum of an equilateral rectangular 
pyramid. 

i / /A -A 2 

1/ S 2 — I — 7) — J = h ; S being the slant height, A a side of the 

greater base, and a a side of the less. 

1/ F 2 — ( — — - J x 2 = h ; F being the slant height meas- 
ured along an angle. 

* Diameter X -7071 = side of inscribed square. See Centres of Surfaces. 



SECTION IV. 

GEOMETRY —PRACTICAL AND ILLUSTRATIVE. 



Geometry is the science that treats of the properties of figured 
space. It is the science of magnitude in general, and comprehends 
the mensuration of solids, surfaces, lines, and their various rela- 
tions. 



DEFINITIONS. 

A Point has position, but not magnitude. 

A Line is length without breadth, and is either Right*, Curved, or 
Mixed. When no particular line is specified, a right line is meant. 

A Right Line is a straight line, or the shortest distance between 
two points. 

A Mixed Line is a right line and curved line united. 

Lines are 'parallel, oblique, perpendicular, or tangentical, one to an- 
other. 

'An Area, surface, superficies, is the space contained within the out- 
line or perimeter of a figure ; it has no thickness, and is estimated 
in the square of some unit of measure, as square inch, square 
yard, &c. 

A Solid has length, breadth, and thickness, and its contents are 
estimated in the cube of some unit of measure. 

An Angle is the diverging of two lines from each other, and is 
right, acute, or obtuse. 

A Right Angle has one line perpendicular to another and resting 
upon it. 

A Triangle, or trigon, is a figure having three sides. 

An Equilateral Triangle has all its sides equal. 

An Isosceles Triangle has two of its sides equal. 

A Scalene Triangle has no two sides equal. 

A Right-angled Triangle has one right angle. 

An Obtuse-angled Triangle has one obtuse angle. 

A n Acute-an-oled Triangle has all its angles acute. 



GEOMETRY. 



173 



A Quadrangle, tetragon, quadrilateral, is a figure having four 
sides. 

A Parallelogram is a quadrilateral figure whose opposite sides are 
parallel and equal. 

A Rectangle is a parallelogram whose opposite sides are equal, its 
angles right angles, and its length greater than its breadth. 

A Square is an equilateral rectangle. 

A Rhomboid is a quadrilateral, having its opposite sides equal and 
parallel, its angles oblique, and a length greater than its breadth. 

A Rhombus, or lozenge, is an equilateral four-sided figure, having 
oblique angles. 

A Trapezium is a quadrilateral having no two sides parallel. 

A Trapezoid is any four-sided figure having two of its sides paral- 
lel, but of unequal lengths. 

A Diagonal is a line joining any two opposite angles of a figure 
having four or more sides. 

A Polygon is a plain figure having more than four sides. 

A Regular Polygon has aH its sides equal. 

An Irregular Polygon has not all its sides equal. 

A Pentagon has five sides ; a hexagon, six ; a heptagon, seven ; 
an octagon, eight ; a nonagon, nine ; a decagon, ten ; an undecagon, 
eleven ; a dodecagon, twelve. 

The Perimeter of a figure is its bounds, limits, or outline. It is to 
other figures what the circumference is to the circle, and the perimeter 
of any portion of a figure is the outline of that portion. 

The Altitude, or height, of a figure, is a perpendicular let fall from 
its vertex, or highest point, to the opposite side or end, its base. 

The Base of a triangle is that side that is placed parallel to the 
horizon ; and of figures in general the base is that end, or side, upon 
which the figure is supposed to stand or rest. The sides of a triangle 
are often called the legs. In a right-angled triangle, the longest side, 
or line which subtends the right angle, is called the hypotenuse, and 
of the other two sides, one is the base, and the other the perpendic- 
ular. 



A Circle is a plane figure, bounded 
by a curve line, called the circumfer- 
ence or periphery, every part of which 
isequi-distantfrom a point within called 
the centre, as A C B D, in the diagram. 
The circumference itself is often called 
a circle. 

The Radius — semi-diameter — is a 
line drawn from the centre to the cir- 
cumference, as O A, or O C. 

The Diameter is a line drawn from 
the circumference through the centre to the opposite side, as A B. 
15* 




174 GEOMETRY. 

A Semicircle is half a circle, or it is half the circumference of a 
circle, as A C B. 

A Quadrant is a quarter of a circle. It is also sometimes a quar- 
ter of the circumference, as A C. 

An Arc is any portion of the circumference, as B c a, or h C g. 

A Chord, or subtense, is a right line joining the extremities of an 
arc, as B a, or h g. 

A Segment is the portion of a circle contained between the arc 
and its chord, as the space between the arc B c a and its chord B a, 
or between the arc h D g, or h C g, and the chord h g. ' 

A Sector is the space between two radii, or lines passing from the 
centre to the circumference, as the space BOfl. 

A Secant is a line that cuts another line. In trigonometry, the 
secant of an arc is a right line drawn from the centre of a circle 
through one end of the arc, and terminated by a tangent drawn 
through the other end ; thus, the secant of the arc B c a is the 
line 6 b. 

A Cosecant is the secant of the complement of an arc, as O b. 

A Sine of an arc is a line drawn from one end of the arc per- 
pendicular to a radius drawn through the other end, as a e, and is 
always equal to half the chord of double the arc ; and the sine of 
an angle is the sine of the arc that measures that angle. 

The Versed Sine is that portion or part of the radius lying be- 
tween the foot of the sine and origin of the arc, as e B, and the 
versed sine of half the arc is that portion of the radius lying be- 
tween the chord and the arc, bisecting or dividing both at their 
centres, as k D, in the arc Ji D g. It is the height of the arc or 
segment. 

The Cosine of an arc, or angle, is that portion of the radius lying 
between the sine and the centre, as e O. 

The Cover sed Sine is the sine of the complement of an arc, or an- 
gle, or the coversed sine of the given arc, or angle ; thus, the line 
a d is the coversed sine of the arc B c a, or of the angle B O a. 

A Tangent is a right line that touches a curve, and which, if pro- 
duced, will not cut it, — the tangent of the arc B c a, is B b. 

A Cotangent is the tangent of the complement of an arc, or the tan- 
gent of an arc which is the complement of another arc to ninety 
degrees ; thus, the cotangent to the arc B c a, is the line C b. 

The Complement is what remains of the quadrant of a circle, after 
the angle has been taken therefrom, — the complement of the arc B 
c a, is a C. 

The Supplement is what remains of a semi-circle after taking an 
angle therefrom, — the supplement of the arc B c a, is a C A. 

A Gnomon is the space included between two similar parallelo- 
grams, one inscribed within the other, and having one angle common 
to them both. 



GEOMETRY. 175 

A Zone is the space between two parallel chords of a circle, — the 
space included between the lines A B and h g. 

A Lune, or Crescent, is the space contained between the inter- 
secting arcs of two eccentric circles, as i n s. 

A Circular Ring is the space between the circumferences of two 
concentric circles. 

An Oval is a figure of an elliptical form made up of arcs of 
circles. 

A Helix is a coil or spiral, or it is part of a spiral line. 

An Ogee, cyma, or talon, is two circle arcs that tangent each 
other, and meet two parallel lines, either tangentical to the lines or 
at right-angles to the lines in given points. 

A Prism is a solid whose bases or ends are any similar, equal plane 
figures, and whose sides are parallelograms. 

A Parallelopiped is a solid having six sides, its angles right-angles, 
and its opposite sides equal. It is a prism, therefore, whose base is 
a parallelogram. 

A Cube is a solid having six equal sides and all its angles right- 
angles. It is a square prism. 

A Prismoid is a solid whose bases are parallel but unequal, and 
whose sides are quadrilateral. 

A Pyramid is a solid having any plane rectilinear figure for its 
base, and all its sides, more or less, terminating in a point, called 
its vertex or summit. 

A Cylinder is a circular solid, having a uniform diameter, and 
equal and parallel circles for its ends. 

A Cone is a solid having a circle for its base, and a true taper 
therefrom to its vertex. 

Conic Sections are the lines formed by the intersections of a 
plane with a conic surface. They are the triangle, circle, ellipse, 
parabola and hyperbola, 

A Conoid is a solid generated by the revolving of a parabola, or 
hyperbola around its axis. 

A Spheroid is a solid generated by the revolving of an ellipse 
about either of its axes or diameters. 

The Transverse or Major axis of an ellipse is its longest diameter, 
or the distance, lengthwise, through its centre. 

The Conjugate or Minor axis of an ellipse is the shorter of the 
two diameters, — a right line bisecting the transverse. If the gen- 
erating ellipse revolves about its major axis, the spheroid is prolate, 
or oblong ; if about its minor axis, it is oblate, or flattened. 

A.n Ordinate is a right line drawn from any point of the curve of 
a conic section to either of its diameters, and perpendicular to that 
diameter. 



176 



GEOMETRY. 



The Abscisses, of a conic section are the parts of either diameter, or 
axis, lying between their respective vertices and an ordinate. 

The Parameter — latus rectum of a parabola — is a third propor- 
tional to any diameter and its conjugate. In the parabola it is a third 
proportional to any abscissa and its ordinate, or to the altitude of 
the figure and half the base. 

The Focus is the point in the axis where the ordinate is equal to 
half the parameter. 

A Sphere, or globe, is a perfectly round substance — a solid con- 
tained under a curved surface, every point of which is equally distant 
from a point within, called the centre. Its axis, or diameter, is any 
right line passing from a side through the centre to the opposite side. 
A hemisphere is half a sphere. 

A Frustum, of any solid figure, as of a cone, pyramid, etc., is the 
part remaining after a segment has been cut off. 

An Ungula is the section of a cylinder cut off by a plane oblique 
to the base. 

The Slant height of a regular figure is the length of one of its 
sides, or the distance from the outline of its base to its vertex, or 
summit. 

To bisect a right line, A B, by a perpendicular. 
Set one foot of the dividers in A, and with the other extended so 
as to reach somewhat beyond the middle of the line, describe arcs 
above and below the line ; then, with one foot of the dividers in B, 
describe arcs crossing the former ; a line drawn from the intersection 
of the arcs above the line to the intersection of those below, will divide 
the line into two equal parts. 
To erect a perpendicular on a given point in a 
straight line, or to draw a line at a right 
angle to another line. -^ 

Set one foot of the dividers in the given / 

point, c, and with the other extended to any ... / 
convenient distance, as to A, mark equal dis- A « ^-.... 
tances on each side c, as c A, c B ; and from 
A and B as centres, with the di\ iders ex- 
tended to a distance somewhat greater than 
that between c and A, or c and B, describe 
arcs cutting each other above the line, as at 

d; a line drawn from the intersection of the arcs, d, to the point c, 
will be perpendicular to the line A B, or will form a right angle with 
the line c A, or c B. 

From a point, d, to let fall a line perpendicular to another line, A B. 
Set one foot of the dividers in d, and with the other extended so as 
to reach beyond the line A B, describe an arc cutting the line A B, 



GEOMETRY. 



177 




in e and n; then with one foot of the dividers in e, and the other ex- 
tended to more than half the distance between e and n, describe the 
arc g; then with one foot of the dividers in n, describe an arc cutting 
the arc g in g ; a line drawn from the point d through c to the inter- 
section of the arcs at g, will be the perpendicular required. 
To erect a perpendicular upon the end of a line, as at a, on the line A c, 

Set one foot of the dividers in c, and, at any convenient radius, de- 
scribe the arc e h k; with one foot of the dividers in e, cut the arc in 
h, and with one foot in h, cut it in k; from A as a centre, and k as a 
centre, describe arcs cutting each other at d; a line drawn from the 
intersection of the arcs, d, to the point c, will be perpendicular to the 
line A c. 

To draw a circle through any three given points not in a straight line, 
and to find the centre or radius of a circle, or arc. 

Let the given points be a, b, c. With the dividers opened to any 
convenient distance, and either point the centre, 
(as b,) describe the portion of a circle r s t, 
and with the same radius and a the centre, / 
describe an arc cutting r s t in r and s, 
and with the same radius and c the centre, 
describe an arc cutting r s t in t and 5 ; 
draw lines through the points where the arcs 
cut each other, (the lines r s and t s,) and their 
point of union will indicate the centre of the 
circle or point, from which as a centre a circle, 
through the three given points. 

To find the length of an arc of a circle. 

Take | the length of the chord of the arc, (A B,) and with the 
dividers at that radius, and A the centre, 
cut the arc in c ; also, with the dividers Jk 

at the same radius and B the centre, cut / \ ~~-~~-^3, 

the chord in d; draw the line c d, and -&. 
twice its length is the leng'th of the arc, nearly. 
From a given point, to draw a tangent to a 
circle. 

Let A represent the given point, and C, the 
centre of the circle. Draw a line from A to 
C ; bisect the line, and with the point of bisec- 
tion as centre, describe the semicircle ADC; 
then draw a right line, A B, cutting the semi- 
circle at the point where it intersects the circle, 
which is the tangent sought. 
To draw from or to the circumference of a circle, lines tending to the 
centre of said circle, when the centre is inaccessible. 

Divide the circle, or such portion thereof as required, into the 



if drawn, will pass 





178 



GEOMETRY. 




desired number of equal parts, and designate the points of division, upon 
or at an uniform distance from the pe- 
riphery ; then, with any radius less 
than two of the parts, and the re- 
spective points as centres, describe 
arcs cutting each other, as A 1, d 1 ; 
c 2, e 2, &c. ; draw the lines c 1, d 2, 
e 3, &c, which tend to the centre, as required. 

When a 'portion of a circle or segment only is used, to draw the end lines, 
A 5, B s. 
Produce the arc each way beyond A and B to a distance equal to 
that between A and c, and with the extremes of the extensions as 
centres, and the second points inward therefrom as centres, describe 
the intersecting arcs, s s. Or, with c as a centre, describe the arc s, 
and with the radius c 1, and A or B as the centre, describe the inter- 
secting arcs ; lines from the intersection of the arcs, s s, to their 
respective points, A and B, will tend to the centre, as required. 

To describe an elliptic arch on a given conjugate diameter. 
Let A B be the given diameter, which bisect, and from the point 
of bisection, e, erect the perpendicular 
ef, equal in length, or proportional in 
length to the height of the intended 
arch ; make e a, eb, each equal to ef, 
and bisect ef in o ; draw the lines a o c 
and bod, and with the radius a B or 
b A, and a and b as centres, describe the 
arcs Ad and Be; then, with the radius 
o d, or o c, and o as the centre, describe the arc dfc, and the arch is 
completed. 

To describe an Ellipse of given length and breadth. 
Let the line A B equal the given length, or transverse diameter, 
and the line C D the conjugate, and let these lines bisect each other, 
forming right angles, on either side, as 
at e. Lay off the distance C D on the 
line A B, as from A to /, and divide 
the distance I B into three equal parts. 
From e, on the line A B, set off two of 
the parts each way, as e a, e h ; and 
from a, or A, designate the distance a k 
on the line C D, as at i and n ; from i 
draw the lines i t and i r, and from n, 
the lines n s and n k, passing through 
the points a and h and cutting each other 





GEOMETRY. 



179 



E ^ 2 3 


D 3 2 ¥ 


N 


^T 


mw 


A. 1 2 3 


C 3 2 1 B 



therein. From the point n, as a centre, describe the arc s k, and 
from i, as a centre, the arc r t; also, from «, as a centre, describe 
the arc t s, and from A, as a centre, the arc& r, and the required ellipse 
is drawn. 

Note. — An architrave of any depth desired, may be readily descrioed on the above. 

To construct an arc or segment of a circle of large radius. 
Draw the chord A B equal in length, or proportional in length, to 
the chord of the arc intended ; also draw E F parallel to the chord and 
at a distance therefrom equal to the 
height of the intended segment. 
Bisect the chord, and from the 
point of bisection erect the perpen- 
dicular C D ; draw the right lines 
AD,DB, and draw A E at a right 
angle to A D, and B F at a right angle to B D, and erect the perpen- 
diculars A /, Bf. Divide A B into any even number of equal parts, 
and divide E F into the same number of equal parts, and draw the 
lines 1, 1 ; 2, 2 ; 3, 3, &c. Divide AfBf each into half the 
number of equal parts the chord A B is divided into, and draw lines 
from D to the points of division, respectively. A curve passing 
through the intersections of the crossing lines bearing the same num- 
ber will describe the arc required. 

To describe an elliptic arch, the span and rise being given. 
Bisect the given chord, or span, A B, witb a line at right angles 
therewith, and let the portion of the line C D be the rise intended. 
Erect A a equal and parallel to C D, 
and draw the line a D equal and parallel 
to A C. Bisect A a in t, and draw the 
line t D ; make C b equal to C D, and draw 
the line a b. Bisect the portion of t D 
lying between the line a b and the point D, 
and from the point of bisection, at right 
angles to t D, draw the line meeting D C 
in g ; then draw the line a g. Let/ des- 
ignate the point where the line a g cuts 
A B, and make h B equal to A f, and draw 
the line g h e. With the radius g D, and ^asa centre, describe the 
arc from the line a g to e, and with the radius A f, and / and h as 
centres, describe the arcs A s, e B, which will complete the arch 
required. 




180 



GEOMETRY. 




To draw a Gothic arch. 

Fig. 1. — With the chord A B as radius, and A and B as centres, 
describe the arcs A c and B c. 

Fig. 2. — Divide the given chord, A B, into three equal parts, and 
with two of the parts as radius, and b and c as centres, describe the 
arcs A d, B d. 

Fig. 3. — Divide the given chord, A B, into three equal parts, e 
and/, and from the points A and B let fall the perpendiculars A a, 
B b, equal in length to two of the divisions of the chord. Draw the 
lines a h and b g, passing through the divisions e and/, and with one 
of the divisions as radius, and e and / as centres, describe the arcs 
A g, B h ; also, with the radius ah, or g b, and a and b as centres, 
describe the arcs g v and v h. 

Fig. 4. — Divide the given chord, A B, 
into three equal parts, a and b, and with the 
radius two of the parts, and A,a,b, and B as 
centres, describe the four arcs b c, B a?, a d, 
Ac; then, with the radius one of the parts, 
and a and b as centres, describe the arcs A h 
and B g; then, with the radius c g, or d h, 
and c and d as centres, describe the arcs h i 
and g i. 




GEOMETRY. 



181 




To describe a Regular Polygon of any number of sides not exceeding 
twelve, to a given face or chord line. 
Let A B be the given face, which bisect, 
and from the point of bisection, and at aright 
angle to A B, draw the line C D. With the 
radius A B, describe the arc B 6, and divide 
the arc into six equal parts, and from 6 con- 
tinue the divisions on the line C D, as v, 7, 
8, 9, &c, to 12. A circle whose radius is 
B v, B 6, B 7, &c, will contain the given 
face or side (A B) 5, 6, 7, &c, times. 

To inscribe any Regular Polygon in a given circle, or to a given 
diameter. 

Let A B be the given diameter, which divide 
into as many equal parts as the polygon is to 
have sides; then, with the radius A B, and A 
and B as centres, describe the arcs cutting each 
other as at C. From the intersection of the 
arcs at C, draw a line through the second point 
of division on the diameter to the periphery, as 
C D, and the chord of the arc, D A, will be one 
side of the polygon required nearly. 

Let a pentagon be required, See Fig. 




To circumscribe a Regular Polygon of any 
given number of sides, about a given circle. 
Divide the given circle into as many equal 
parts as the polygon is to have sides, and de- 
fine the points of division on the circle ; then 
draw lines from the centre, o, to each of the 
respective points, as o a, o b, &c. Through 
these points, and at a right angle to the line 
leading from the centre thereto, draw the 
lines A B, B C, &c, which will complete the figure. 

Let it be required to circumscribe with a pentagon ; — see Fig. 

To construct a square whose area shall be that of a given triangle. 
Let A B C be the given triangle. 
From B let fall the perpendicular B a; 
make A b equal to half B a ; bisect b C, and 
from the point of bisection as a centre, de- 
scribe the semicircle b d C : erect the perpen- 
dicular, A d, which will be the side of the 
square required. 

16 




182 



GEOMETRY. 



3 S 12 3l) 5 6 rjf 




To construct a Parabola 
Let A B equal the base, which bisect, 
and upon the point of bisection erect the 
perpendicular C D, the altitude; let the 
line E F be half the length of A B, and 
let it lie parallel to A B, and at the dis- 
tance D C from the base, and let it be bi- 
sected by C D, as at D. Draw the lines 
E A, F B, and divide them, together with 
E D, D F, into four or any number of 
equal parts, as 1 1,2 2, &c, and draw the 
lines connecting the respective points of 
division. The inner intersections of the 
said lines with each other define the curve of a parabola. 

To construct a Hyperbola. 
Let A B equal the longest or transverse diameter, and C D, 
pendicular to it, the 
conjugate, and let the 
line A B be produced or 
extended from its re- 
spective limits each 
way, as to a, r, r, &c. 
Bisect A B in o, and 
with the radius o C, or 
o D, and o as the centre 
describe the circle C 
D a. Divide A B pro- 
duced from B, into any 
number of parts, as r, r, 
r, &c, and with the ra- 
dii A r and B r, and the 
foci a and e as centres, 
describe arcs cutting 
each other as in s s, &c. The intersections of the arcs with each other 
will define the curve of the hyperbola. 




To bisect any given triangle. 
Let A B C be the given triangle. 
Bisect one of the legs, as A B, and with the 
point of bisection as a centre describe the semicir- 
cle, B a A ; bisect the semicircle, as at a, and with 
the radius B a, and B as a centre, describe the 
arc a b ; from b erect the line b c parallel to A C, 
which will bisect the given triangle, or divide it 
into two equal parts, as required. 




GEOXETKY. 



183 



To draw an equilateral triangle whose area shall 
be that of two given equilateral triangles. 
Let the given triangles be A and B. 
Draw a line D E equal in length to one side of 
the larger triangle, and upon one end thereof, and 
at a right angle therewith, erect a line equal in 
length to one side of the less triangle, as DC; 
then draw the line C E, which will be one leg of 
the triangle required. The equilateral triangle 
C E F contains an area equal to the triangles A 
and B. 

Note. — The same process is applicable to rectangular figures. 





To draw a circle whose area shall he that of two given circles. 
Let the circles A and B be the given 
circles. Draw a line whose length shall 
be equal to the diameter of the larger 
circle, and upon the end thereof erect a 
perpendicular equal in length to the di- 
ameter of the lesser circle, as C D E ; 
draw the line E C, and bisect it in i, and 
with the radius i 0, or i E, and i as the 
centre, describe the circle E D F, whose 
area will be that of the two given circles. 

To construct a toothed or cog wheel. 
Divide the pitch circle, a a, into as many equal parts as there are 
to be teeth or cogs ; then, 
with the dividers extended to 
l| times one of those parts, 
and the point s as a centre, 
describe the arcs i v, v i ; 
then, with the same radius, 
and r and t as centres, de- 
scribe the arcs vf and/u, so 
continuing until the upper 
sections of all the cogs are defined. The lower sections are bounded 
by straight lines tending to the centre of the wheel. See Teeth of 
Wheels. 

Note. — The pitch of a wheel is the rectilinear distance from the centre of one cog to 
the centre of the next contiguous, measured upon the pitch circle ; and that portion of 
the length of a tooth lying between the lines a and 6 is usually made equal to i the pitch; 
and that portion lying between the lines a and c is usually made equal to ± the pitch. 




184 CONIC SECTIONS. 



OF THE CONIC SECTIONS. 

The conic sections are the elements of geometry. They 
are lines, and nothing more. The doctrine of their relation? is the 
science of geometry. Geometry is lines in position. 

The conic sections are the lines formed by the intersections of a 
plane with a conic surface. They are the triangle, circle, ellipse, 
parabola, and hyperbola. With a single exception, they are curved 
lines or curves. 

To the conic sections belong the foci, parameters, abscissce, and 
ordinates, which are explained in a general way under defini- 
tions, p. 175, 176. 

The locus, or place, of a conic section is determined by a point 
called the generatrix, which is supposed to move in accordance with 
the law of the line. 

The equation of a line expresses the relation between the ordi- 
nate and abscissa of every point in the line, or between the co-ordi- 
nates of every point in the line. 

The parameter is a double ordinate, and passes through the focal 
point. 

The summit of an ordinate is a point in the locus of a line, in 
the path of the generatrix. Thus by the plotting of ordinates or 
double ordinates to different abscissae, the line is defined ; and, by 
properly connecting the points, it is practically constructed or 
formed. 

The eccentricity of a conic section is its deviation from the centre. 

The axis of abscissce is the axis, or diameter, in which the abscissae 
are taken, or the axis, or diameter, that is divided into abscissae. 

The axis of ordinates is the axis, or diameter, that is parallel to 
the ordinates. 

The origin of a conic section is in the summit of the axis of 



The axis of abscissae proper is the major axis in any conic 
section. 

The asymptote is a right line that continually approaches a curve, 
but never meets it, however far both may be extended (see Con- 
structions, Hyperbola, p. 182). 

In all the conic sections, the parameter is a third proportional to 
any diameter and its conjugate. 

The radius vector, or vector, is a right line joining the centre of 
the sun to the centre of a planet. It is an element in astronomy, 
and has one of its extremities in the focus. 



LINES AND SUPERFICES. 185 



LONGIMETRY AND PLANIMETRY; 

OR, 

LINES AND SUPERFICIES. 



TRIANGLES. 

A triangle is the simplest form in geometry, and the most im- 
portant. It is a plane, three-sided, rectilinear figure, or is made up 
of three straight lines. It is the measure chiefly of itself, and is 
the measure of almost every geometrical form or structure. It is 
the measure of the resultant of forces acting from different direc- 
tions upon the same body. Similar triangles are measures of each 
other. 

It is generated, when a plane intersects both sides of a cone, 
from its apex along the plane of its axis. 

As a conic. section, it is simply two diverging lines having their 
origin in the same point, and subtended by an ordinate. 

Its equation is expressed by x \ y \ ; x' '. y 1 ; x being any abscissa 
on either leg reckoned from the angle opposite the ordinate, y the 
ordmate, x' any other abscissa on the same leg reckoned from the 
same angle, y' ordinate to abscissa x'. 

In trigonometry, it is a figure having three sides and three 
angles, and is supposed to have one of its angles in the centre of a 
circle, of which one of the sides is radius. 

In geometry, it is a figure having three sides, three angles, an 
apothegm, or perpendicular, and an area. 

Under its most favorable form (equilateral), it contains less area 
than a square of the same length of perimeter by 23 per cent. ; 
and less than a circle of the same length of perimeter by nearly 
40 per cent. 

16* 



186 



LINES AND SUPERFICIES. 




Right-angled Triangle : — ADC, diagram. 

The longest side of this figure, 
A C, is usually called the hy- 
potenuse, and the other two 
sides, A D and D C, are called 
the sides or legs. The legs are 
perpendicular one to the other, 
and form the right-angle, D, or 
angle of 90°. 

If one of the legs be made 
base, the other will be perpendicular, and will be the altitude of the 
figure ; for the altitude of any triangle is a perpendicular, dropped 
from the vertical angle to the opposite side or base, and either side 
may be made base ; thus, if A D be made base, D C will be perpendic- 
ular, and will be the altitude of the figure, and if D C be made base, 
A D will be perpendicular, and will be the altitude of the figure. 

If the hypotenuse be made base, the legs, A D and D C, will still 
be perpendicular one to the other, but the altitude of the figure, D E 
or E D, a perpendicular to the hypotenuse, will not be shown. 

Whether the legs have equal lengths or unequal, so far as regards 
the principles of the figure, is immaterial. 

/v/(AD 2 -fDC 2 )=AC; V ( AC 2 — DC 2 ) = A D ; 

V(FC 2 -AD 2 ) = DC. 

A~I) 2 _i_ A C = A E } Converting the right-angled trian- 

/ 9 - — o ' > e;le A D C into two right-angled tri- 

V(A D:-AE)=ED.i Ingles, A E D and C E D, E D a 
leg common to both , and perpendicular to A C , and the altitude of 
the triangle ADC, therefore, A C being base. 



DC>AC = CE. 
V(D~C 2 — Cf) =ED. 
AD 2 4- A E = A C. 
"DC 2 -h C E = A C. 
ED 2 -i- A E = C E. 



A C X A E = A D~. 
ACX CE=ITC 2 . 
A E X C E = ED 2 . 
AD 2 : AC 2 : : A E : A C. 
D~C 2 : AD 2 : : C E : A E. 



D C : AC" ::CE : AC. 

Twice the area of a right-angled triangle, divided by the hypo- 
tenuse, is equal the perpendicular to the hypotenuse, and the 
respective triangle will thereby be divided into two right-angled 
triangles having a side common to both. 



LINES AND SUPERFICIES. 



187 



Half the product of the two legs of a right-angled triangle equals 
the area of that triangle. 

2 area 2 area 

perp. to hypot. = hypotenuse. g— -^ = required leg. 

Of Oblique-angled Triangles. 

Every triangle not a right-angled triangle is either an acute- 
angled triangle or an obtuse-angled triangle, and these two (the 
acute-angled and obtuse-angled) are classed under the general name 
oblique-angled. The following principles are alike applicable to 
either. 



Let A B C be the triangle. 




AC' 



BC 2 



2AB +4AB-AD, 

distance along the base, from 
the angle formed by the base 
and longest vertical side, at 
which a perpendicular dropped 
from the vertical angle will fall ; 
and 

V (AC 2 — AD 2 ) =DC, the 
perpendicular alluded to ; thus dividing the obtuse-angled triangle 
ABC into two right-angled triangles, ADC and B D C, D C a leg 
common to both. 



Or, a/ (A C + A D X A C — A D) = D C ; for the sum of any 
two quantities multiplied by their difference is equal to the differ- 
ence of their squares. 

AB '-^ g! + £ B C = B g , and V ( AB 2 - B^ 2 ) - A g , 
perpendicular to B C produced. 



AB-BC 2 



AC 



2 A C '2 

pendicular to A C produced 



A h, and V(A B 2 - A h 2 ) = B h, per- 



AC=AB+BC- 2ABXBD 



BD = 



_AB + BC— AC 



2AB 



188 LINES AND SUPERFICIES. 

ac+Fb-b"c 

AD — 2pj * 

AC- A^ 2 =c"^andC(7-[-B C = B<7; Ag -\-Cg=AC. 

Twice the area of any triangle divided by the base is equal the 
perpendicular to the base. 

Half the side of any triangle multiplied by the perpendicular 
to that side is equal the area. 

Note. — In an obtuse-angled triangle, a perpendicular dropped from either 
of the acute angles will fall outside the figure. 

In an equilateral triangle, a perpendicular dropped from either angle will 
bisect the side opposite; and the triangle will thereby be divided into two 
equal and similar right-augled scalene triangles. 

In an isosceles triangle, a perpendicular dropped from the angle included 
by the equal sides will bisect the side opposite, and the triangle will thereby 
be divided into two equal and similar right-angled scalene triangles, or two 
equal and similar right-angled isosceles triangles : dropped from an angle 
opposite one of the equal sides, it will fall outside the figure if the triangle 
be obtuse ; inside, if it be acute. 

RECAPITULATIONS. 

To find the Perpendicular of an Oblique-angled Scalene Triangle, 
the Sides being given. 

Let I represent the longest side, s the shortest side, m the inter- 
mediate side, and h the perpendicular. 

When I is made base (B), 

h= I* 2 — ( b2 + * 2 — m '\ 2 = L 3 — ( V-L-m 2 — g 2 Y 

When s is made base, / n , -q 2 

h 



When m is made base, 

h = 



J ,.-(a^-) : 



To find the Area of a Triangle. 
Rule. — Multiply the base by half the perpendicular height, or, 
the perpendicular height by half the base ; or, multiply the base 
by the altitude, and divide the product by 2, and the quotient will 
be the area. 

Example. — The base A C, of the triangle A C B, 
is 12 feet, and the altitude, D B, is 4 feet and 6 inches ; 
required the area. ; 

4.5 X ~Y= 27 square feet. Ans. ad c 

To find the Area of a Triangle by Means of the Sides. 
-Rule. — Add the three sides together, and from half the sum sub- 
tract each side separately ; multiply the half sum and the three re- 



LINES AND SUPERFICIES. 189 

mainders into each other, and from the product extract the square 
root, which will be the area sought. 

Example. — The sides of a~triangle are 30, 40, and 60 rods; 
what area has the triangle ? 

30 -f- 40 -}- 60 H- 2 = 65 = a sum of the three sides. 
65 — 30 = 35, first remainder. 
65 — 40 = 25, second remainder. 
65 — 60 = 5, third remainder. 
65 X 35 X 25 X 5 = ^284375 = 533.26 square rods. Ans. 

To find the hypotenuse of a triangle, the other two sides being given. 

Rule. — Add the square of the base and square of the perpen- 
dicular together, and the square root of the sum will be the 
hypotenuse. 

Example. — The distance from the base of a building 
to the sill of an attic window — perpendicular of the 
triangle, B C — is 40 feet; what must be the length of 
a ladder, — hypotenuse of the triangle, AC, — placed 
on a level with the base of the building, and 12 feet 
therefrom, — base of the triangle, A B, — to reach to 
the sill of said window ? 

40 2 -j- 12 2 = ^1744 = 41 f feet. Ans. 

The hypotenuse and one of the sides of a right-angled triangle being 
given, to find the length of the other side. 
Rule. — Add the hypotenuse and the given leg together, mul- 
tiply the sum by the difference of their lengths, and the square root 
of the product will be the side sought. 

• Example. — The sill of a window in a building standing on the 
edge of a stream is 30 feet above the water, and a line which has 
been extended therefrom directly across the stream to the oppose 
shore is found to measure 80 feet ; required the width of the stream 
at that place. 
80 — 30 = 50, and (80 -f- 30 )X 50= y/5500 = 74.16 feet. Ans. 

To find the height of an inaccessible object, C, or the length of the 
perpendicular B C. 
Rule. — Upon a plane, at a right angle q 

to the base of the perpendicular, as A B, 
erect, at any convenient distance from the 
base, a perpendicular staff D E, and con- 
struct the hypotenuse A E in the direction 
A C ; then, as the base AD is to the per- 
pendicular staff D E, so is the base A B to 




the height B C. AD 



190 



MENSURATION OF SUPERFICES. 



Example. — The perpendicular D E, of the right-angled triangle 
ADE, being 3 feet, and the base, AD, 5 feet, what is the perpen- 
dicular of a similar triangle, ABO, whose base, A B, is 100 feet? 
5 : 3 :: 100 = 60 feet. Ans. 



To find the distance from a given point, B, to an inaccessible object, 
as at A. 
Rule. — Draw a line B in the direction A, 
and from the point B, at a right angle to B A, 
draw, to any convenient length, a line B C ; form 
a right angle to B C, from the point C, and con- 
vert into a triangle by carrying the hypotenuse, 
D E, to the line B C, in the direction D A ; then, 
as the distance C E is to the distance C D, so is 
the distance B E to the distance B A. 

Example. — The distance from C to E is 4 
rods, the distance from C to D, 15 rods, and the B 
distance fromE to B, 22 rods, whereby required the distance from B 
to A. 

4 : 15 :: 22 = 83| rods. Ans. 





OF QUADRILATERAL FIGURES. 




F G 


E|j|||jgjF 


frngf 


G^ — US Ih 


u i 


Rectang-le. 


Rhombus. 



Square. 

To find the area of either the above figures. 

Rule. — Multiply the length by the breadth or perpendicula) 
height, and the product is the area. 

Example. — How many square feet in a floor whose form is 3 
rhomb, or rhombus, each side being 12 feet, and the perpendicu- 
lar or right-angular distance from the side F G to the side H I being 
8 feet ? 

12 X 8 = 96 square feet. Ans. 



LINES AND SUPERFICIES. 191 

The above figures are 2^ ara ^ e ^°9 rams -> f° r a parallelogram is a 
quadrilateral whose opposite sides are equal. 

A square is an equilateral rectangular parallelogram : it is two 
equal and similar right-angled isosceles triangles by either of its 
diagonals ; draw both its diagonals, and it will be made up of four 
equal and similar right-angled isosceles triangles. 

A rectangle is a right-angled parallelogram that has more length 
than breadth : it is two equal and similar right-angled scalene tri- 
angles by either of its diagonals ; by both, it is two equal and 
similar obtuse-angled isosceles triangles, and two equal and similar 
acute-angled isosceles or equilateral triangles. 

A rhombus, rhomb, or lozenge is an equilateral oblique-angled 
parallelogram : it is two equal and similar obtuse-angled isosceles 
triangles by its longest diagonal ; it is two equal and similar acute- 
angled isosceles or equilateral triangles by its shortest diagonal; 
it is four equal and similar right-angled scalene triangles by both 
its diagonals. 

A rhomboid is an oblique-angled parallelogram that has more 
length than breadth : it is two equal and similar right-angled sca- 
lene triangles by either of its diagonals ; by both, it is two equal 
and similar obtuse-angled scalene triangles, and two equal and 
similar right-angled scalene triangles. 

To find the Area of a Trapezoid. 

Rule. — Multiply the sum of the two parallel sides by the per- 
pendicular distance between them, and divide the product by 2 ; 
the quotient will be the area. a b 

Example. — The side A B, of the trapezoid A B- 
C D, is 48^ feet, the side C D is 72^ feet, and the 
perpendicular distance between the sides is 40£ feet ; 
how many feet area has the figure ? c d 

72 T ^ -}- 48y^ = 1 20f X 40£ = 4890f -J- 2 = 2445 ft. 2£ in. Am. 

To find the Area of a Trapezium. 
Rule — Draw a diagonal through the figure, which will divide 
it into two triangles, and multiply the length of the diagonal by 
half the sum of the altitudes of the triangles ; the product will be 
the area. 

Example. — The diagonal A D, in the trapezium /T*^. 

A B C D, is 54 rods in length, and the altitudes of A y%_\ ^^ 
the triangles formed by the introduction of the Y~j ^^^ 
diagonal are 20 rods and 26 rods ; required the \^^ 
area of the figure. c 

26 -f- 20 -f- 2 = 23 X 54 = 1242 square rods. Ans. 





192 MENSURATION OF SUPERFICIES. 

Having ihz figure of a rhombus, rhomboid, or trapezoid, the perpen- 
dicular, whereby to find the area, may be found by the following 
method. 

Suppose the diagram AB C D. 

A D and E C sides parallel to B _^3 

each other. 

A D base. 

A B side inclining to the base 
and whose length is known. 

Then, at any convenient dis- 
tance from the angle A, on A D, ^/^ ' 
erect a perpendicular to AD that Jl 
will cut the side A B, as i h ; then will A i be to i h as A B to B E, 
perpendicular required ; or A i will be to A h as A B to A E, dis- 
tance from the angle A, on A D, at which a perpendicular dropped 
from B will fall ; and 

/V (A B 2 — A E 2 ) =BE, perpendicular required. 

Example. — The figure A B C D is that of a field whose side A B 
is 68 rods : Afis 4.2 feet, and i h is 3.7 feet ; required the perpen- 
dicular distance from the side AD to the side B C. 

4.2 : 3.7 : : 68 : 59.9 rods. Ans. 

Example. — A bin in the form of a trapezoid, A B C D (Fig.), has 
a side A B inclining to AD that is 12 feet in length, and a perpen- 
dicular to A D, erected on A D, that cuts the side A B, gives the 
segment A i 2 feet, and the segment AH| feet ; required the per- 
pendicular distance from the side A D to the side B C. 

2 : 1.5 : : 12 : 9, and V(12 2 — 9 2 ) = 7.94 feet. Ans. 
A i : A B : : A h : A E, diagram. 
A E : A B : : A h : A i, 
A h : E h : : A i : B i] « 
i h : A i : : B E : AB, » 

A h : ih : : AE : BE, " 

To find a diagonal of the above figure A B C D. 

V (BE 2 -f E~D 2 ) = B D, diagonal. 

Having the area of a rhombus, rhomboid, or trapezoid, and the sum oj 
any two sides of the figure that are parallel to each other, the perpen- 
dicular to those sides will be found, if we divide twice the area by the 
sum referred to. 

Suppose A D and B C the given sides ; 

B E the perpendicular required ; then 



MENSURATION OF SUPERFICIES. 193 



A area 


area 


:AD + BC. 


AD-f-BC-^' 


PE~ 


area 
-rm=, — A D = B 0. 


2 area 


BC — AD. 



If through any four-sided figure a diagonal he drawn or be supposed 
to be drawn, the figure will be converted into two triangles, each of 
which will have a side, the diagonal, that will be common to them 
both ; and the length of that side or diagonal, which is an indispen- 
sable element in calculating the area of a trapezium, may be found, 
when more simple means may not be resorted to, by one or the other 
of the following methods : — 

Suppose the figure represented by the diagram A B C D. 

B C3 to A, diagonal required. 

T*\ .&/ Construct a partial diagonal, Gt, 

^-''*'* \/ direction C A, and thereon erect a 

perpendicular to C t that will cut 





an adjacent side , as r t cutting the 
side 6 D ; then C r I C t I I C D : 
F, andO r I r t I : C D : D F ; and 
V (A D 2 - DF 2 ) = A F, and C F 
-f- A F = A, diagonal sought. 

Example. — A structure in the form of a trapezium, A B C D, 
(Fig.) has r 8 inches, C t 6 inches, and r t 5 inches; the side 
C D is 16 feet, and the side A D is 20 feet ; required the length of 
a diagonal C A. 

8 : 6 : : 16 : 12 feet, C F, or distance from C, on a line C A, at 
which a perpendicular dropped from D will fall, and 

8 : 5 : : 16 : 10 feet, D F, or length of that perpendicular, then 

20 2 — 10 2 = V300 = 17.32 + 12 = 29.32 feet, length of a diag- 
onal A, or of the side A C of the triangle A B, or of the side 
C A of the triangle CAD. 

Again, suppose the same diagram, the side A D accessible, and 
D to B the diagonal required. 

Dn:DE::Ds:DB, diagonal sought. 
17 



194 



MENSURATION OF SUPERFICIES. 



OF POLYGONS. 
To find the area of a regular polygon. 

Rule. — Multiply the length of a side by half 
the distance from the side to the centre, and that 
product by the number of sides; the last product 
will be the area of the figure. 

Example. — The side A B of a regular hexagon 
is 12 inches, and the distance therefrom to the 
centre of the figure, d c, is 10 inches ; required 
the area of the hexagon. 

-V°- X 12 X 6 = 360 sq. in. = 2± sq. feet 

Table of angles relative to the construction of Regular Polygons with 
the aid of the Sector, and of co-efficients to facilitate their construction 
without it; also, of co-efficients to aid in finding the area of the figure 
the side only being given. 




Names. 


No. 

of 

sides. 


Angle 

at 
centre. 


Angle 

at 

circum. 


Perp'n. 

side 
being 1. 


Length of 

side, radius 

being 1. 


Radius of 

circle, side 

being 1. 


Radius of 

circ. perp. 

being 1. 


Area, 

side 

being 1. 


Triangle, 
Square, 
Pentagon, 
Hexagon, 


3 

4 
5 
6 


120O 
90 
72 
60' 


60° 

90 
108 
120 


0.28863 
0.5 

0.6882 
0.866 


1.73205 
1.4142 
1.1755 
1. 


.5773 
.7071 
.8506 
1. 


2. 

1.4142 
1.236 
1.155 


0.433013 
1. 

1.720477 
2.598076 


Heptagon, 


7 


5lf 

45 
40 
36 


128* 
135 
140 
144 


1.0382 


.8677 


1.152 


1.11 


3.633912 


Octagon, 
Nonagon, 
Decagon, 


8 
9 
10 


1.2071 
1.3737 
1.5338 


.7654 

.684 

.618 


1.3065 
1.4619 
1.618 


1.0823 

1.06 

1.05 


4.828427 
6.181824 
7.694209 


Undecagon, 


11 


32 TT 


147 T 3 T 
150 


1.7028 


.5634 


1.7747 


1.04 


9.36564 


Dodecagon, 


12 


30 


1.866 


.5176 


1.9318 


1.035 


11,196152 



Note. — " Angle at centre " means the angle of radii, passing from the centre to the 
circumference, or corners of the figure. 

" Angle at circumference " means the angle which any two adjoining sides make with 
each other. 

Every circle contains its own radius, as a chord line, exactly six 
times ; therefore, 

To describe a polygon with the aid of the sector. 

Rule. — Take the chord of 60° on the sector, and describe a circle ; 
then, with the chord, (on the same line of the sector,) of as many 
degrees as indicated in the table, for the respective polygon — col- 
umn, " angle at centre " — space off the circle, and each space will 
be the side of the polygon required. Thus, for a decagon, take the 
chord of 60° on the sector for the radius of the circle, and the chord 
of 36° on the same line of the sectors for a side. 



MENSURATION OF SUPERFICES. 195 

Example. — 1. The radius of a circle is 7 feet ; required the side 
©f the greatest regular octagon that may be inscribed therein. 
7 X -7653 = 5.3571 feet, or 5 ft. 4| in., nearly. Ans. 

2. The sides of a heptagon are to be, each, 5 inches ; required the 
radius of circumscribing circle. 

1.152 X 5 = 5.760 = 5| -|- in. Ans. 

3. Each side of a hexagon is 12 inches ; required the distance 
(d c) from the centre of a side to the centre of the figure. 

12 X -866 = 10.392 inches. Ans. 

4. The side of an equilateral triangle is 12 inches; required its 
altitude, — the perpendicular. 

12 X 3 X .28868 = 10.392. Ans. 

5. A Perpendicular, from the centre to either side of a hexagon, 
is required to be 12 inches ; what must be the radius of circumscribing 
circle 1 

12 X 1-156 = 13.872 inches. Ans. 

To find the area of a regular polygon, when the side only is given. 

Rule. — Multiply the square of the side by the number or factor 
in the table — (column, Area) — opposite the name of the respective 
polygon, and the product will be the area. 

Example. — Each side of a nonagon is 6£ rods ; required its area. 
6.5 2 X 6.181824 = 261.18 -j- square rods. Ans. 

To find the area of an irregular polygon. 
Rule. — Divide the figure into trapeziums and trian- 
gles, by drawing diagonals, and find the area of each, 
separately ; the sum of the several areas will equal the 
area of the figure. 

Example. — The outline of the above figure defines an irregular 
polygon ; the enclosed lines divide it into three triangles ; and the 
areas of the several triangles, taken collectively, are equal to, or con- 
stitute, the area sought. To find the areas of the triangles, see Tri- 
angles — Mensuration of. 




196 LINES AND SUPERFICIES. 



CIRCLE. 

A circle is an endless line equidistant in all its parts from a 
point within called the focus, or centre. 

It is generated when a plane revolves about a conic surface per- 
pendicular to the axis, or by a plane cutting through both sides of 
a cone parallel to the base. 

It is equated by means of the triangle, as we shall see. 

Either semi-diameter of two diameters drawn at right angles to 
each other in a circle may be an ordinate. 

In Trigonometry, a circle is the measure of angles, and is sup- 
posed to be divided into parts called degrees, minutes, and seconds ; 
also into semicircles, quadrants, and arcs. In Trigonometry, 
arcs are the measures of angles, and angles are the measures of arcs 
and sides (see Trigonometry). 

In Geometry, a circle-plane is usually called a circle, and its 
boundary line, above described, is called the circumference, or 
periphery ; thus triangles, as they relate to the circle, and are 
drawn within its circumference, may be counted sectional lines of 
the circle, and treated as such. 

Generally, then, a circle is a plane figure having a circumfer- 
ence, a radius, a diameter, an area, &c. 

The ratio of the circumference to the diameter of a circle is 
commonly expressed by the Greek letter it {pi), and tt will be used 
with that signification and no other in this work. The exact value 
of 7T, or the length of the circumference of a circle when the 
diameter is 1, it is well known has never been found: it can no 
more be written with numbers, probably, than can the exact root 
of a surd number. 3.1415926535897932384626433832795028841- 
9716939937. . . =zir. In practice, k is usually taken = 3.1416. 

A circle contains a greater area than any other figure of the 
game length of perimeter or outline. 




MENSURATION OF SUPERFICIES. 197 



OF THE CIRCLE AND ITS SECTIONS. 

Relationship of the sectional lines of the circle, 

one with another, whereby, any two being 

given, any other may be found. 

To furnish several examples, and to exhibit 
the proof with the operation, suppose the lines 
a b and p d given, and let a 6 = 20, inches, 
feet, yards — any linear measure, — and let 
pc?=3.51. 
ab a p 

20 -T- 2 = 10, sine of half the arc, or half the chord of the segment. 

af p d 2 p d 2 c a 

100 -f- 12.32 -T- 3.51 == 32, diameter. 

2ca c a 

32 -i- 2 = 16, radius. 

c a p d c p 

16 — 3.51 = 12.49, cosine of half the arc. 

ca c p p d 

16 — 12.49= 3.51, versed sine of half the arc, or height of segment. 

cp p d ca 

12.49 + 3.51 = 16, radius. 

a p l pd 2 ad 2 ad 2 ad 

100 + 12.32 = 112.32, and V 112.32^ 10.5981, ch'd of half the arc. 

ca 2 ap 2 cp 2 cp 2 cp 

256 — 100 = 156, and V156 = 12.49, cosine. 

ca 2 cp 2 ap 2 ap 2 ap 

256 — 156 = 100, and V100 = 10, half the chord of seg. 

cp 2 ap 2 ca 2 ca 2 ca 

156 -j- 100 = 256, and V256 = 16, radius. 

Radius and height of segment given, to find chord of segment, 
ca — pd = cp,andca 2 — c p 2 = \/a p 2 = «jo,and a p X2 = 2 ap=» 
chord, a b. Ans. 

Chord and height of segment given, to find diameter. 
£ a b -f- V d 2 = ad 2 , and ad 2 -$-pd=2ca. Ans. 

Radius and height of segment given, to find chord of half the arc. 
ca — pd=* cp, and ca 2 — cp?=*ap 2 , and ap 2 -f- p d 2 =s/ad 2 =ad 

Ans. 
17* 



198 LINES AND SUPERFICIES. 

Radius and sine given, to find versed sine, or height. 
£(*' — ap 2 = cp 2 , and ca — cp = pd. Ans. 

Radius and cosine given, to find chord of half the arc. 
ca' -- cp 2 -= ap 2 ,ca — cp—p d,and'apP -\-~p~d 2 /^a~d 2 = ad. 

Ans. 
^Radius and chord of half the arc given, to find sine of half the arc. 
ad- -f- 2 ca=pd, and a d 2 — pd 2 = /^/ap 2 — a p. Ans. 
_Chord of half the arc and sine of half the arc given, to find radius, 
ad- — ap 2 — pd 2 , and ad 2 -r-2pd=ca. Ans. 

Chord of half the arc and sine of half the arc given, to find cosine, 
ad j— ap 2 = pd 2 , and a d 2 -f- 2p d = c a, and ~ca 2 — a~p 2 — 
a/c ir = cp. Ans. 

Radius and sine of half the arc given, to find versed sine, 
c «- — ap 2 = cp 2 , and ca — cp = pd. Ans. 

Sine and Versed Sine given, to find Cosine. 

ap -f- p d" ■=. a d and ad 2 -^-2pd=c a ; ca — pd=cp. Ans. 

The equation of the circle is commonly written y 2 =a 2 — x 2 , 

which, referred to the foregoing, is equivalent to the expression 

ap l =ac — cp", and denotes the same quantities. But, in An- 
alytical Geometry generally, sines and cosines are not known : they 
give place to ordinates and abscissce. Thus in the equation of the 
circle, and in constructing it by plotting the double ordinates of 
different abscissae, ap is an ordinate and cp is its abscissa. Re- 
ferring to the diagram, ap — 2 ac X p d — p d , and this is an 
expression for the equation of the circle : it may be written 
y = ^/(dv — v 2 ). Both c p and dp are abscissas to the ordinate 
a p. The equation of a curve expresses the relation between the 
ordinate and corresponding abscissa of every point in the curve. 

To find the Diameter, Circumference, and Area of a Circle. 

Let d represent the diameter, c the circumference, and A the 
area. 

d = c-r-TTzrc -7- 3.1416 =y/(4A -^-vr) = 1.12838^. 
c = 7rd = dX 3.1416 =7ry/(4A-^7r) = 3.5449i/A. 
A=7rd 2 -7-4=d 2 X .7854 = c^~-4 = c 2 -r-47r=7rr 2 . 



LINES AND SUPERFICIES. 199 

Example. — A line (the chord a b, diagram) was stretched 
across a circular railway 46 feet, and the versed sine, p d, was 
found to be 1.8 feet ; required the diameter, circumference, and 
area of the circle. 

46 -i- 2 = 23, and (23 2 -f- 1.8 2 ) -^ 1.8 = 295.69 feet, diameter. 

Am. 
295.69 X 3.1416 = 928.94 feet, circumference. An&. 

• 295.69 2 X 0.7854, or ^ 68669.55 square feet, or 

> = Y Ans. 

928.94 295.69 j 68669.55 -f- 43560 = 1.576 acres 

To find the Length of an Arc of a Circle. 

Let r represent the radius of the arc ; n, the number of degrees 
in the arc ; d, the diameter of the circle to which the arc belongs ; 
v, the versed sine of half the arc ; and I, the length of the arc. 

, — / v , 3v 2 , 3.5.v 3 3.5. 7.v 4 , \ 

\ ~2.3.d 2.4.5.d 2 2.4.6. 7.of 2.4.6.8.9.d 4 / 

l = itrn~ 180 = .01 745329252m. 

These rules are applicable to all arcs of circles, whether greater 
or less than half the circumference. But the former is too te- 
dious for general practice, and it is not often that we can know 
the number of degees in the arc without the aid of trigonometri- 
cal tables. We can approximate the length with certainty and 
considerable precision by other methods. 

Let b represent the chord of the arc, a b, diagram ; r, the radius 
of the arc ; c, the cosine of half the arc of the salient angle, cp y 
diagram ; Z, the length of the arc of the salient angle ; and 

fc = 0±iV?)l£= 86.13961: then 

*V 2 

When c is equal to or greater than -|&, 
I = 1.5034198&r-^ (r-\- \c + c -^- ) very nearly, 
and n = kb -f- (r -f- \c -4- — ~-^p\ very nearly ; 



/ 86.1396 lb \ 

\r + i c + I3») 



"180 ' ' — Wi r-r-rr-r^rr, 1 verv nearl 7- 



200 LINES AND SUPERFICIES. 

When c is less than -|&, 



/ 0.957107c 



957107c \ 



2kc 
and n = 180 — very nearly ; 

r + if + ^ C 

••• fa I^( 180 - r + ib + i^ ) ^ nCarIy - 

But 27ir = circumference of circle containing 360°. 
The length of the greater arc, or arc of the re-entrant angle, 
therefore = 2itr — I, and contains 360 — n degrees, . • . 

— 7rr(360 — n)-±- 180. 
Now, by the foregoing expressions for n, we obtain the calcu- 
lated angles B-^(r + l C + c -^) , and 2kc -f- (r + \ & + ^- c ) , 

strictly correct for angles 45°, 90°, 135° 7 and 180°; and the errors 
are equal and reciprocal for angles and their supplements of 180°. 
They are all — from to 45°, and from 90° to 135°, and -j- from 
45° to 90°, and from 135° to 180°. The maximum error is only 
2'. It obtains when c and -|& are to each other as 3 to 4, or at 
73° 44' 22" and 106° 15' 38". The mean error (one minute) oc- 
curs at 16° 15' 38", 28° 44' 22", 57° 28' 44", 86° 13' 6", 93° 46' * 
54", 122° 31' 16", 151° 15' 38", and 163° 44' 22". 

Note. — The popular expression for the length of the arc of a circle, when 
the ratio of the arc to the circumference is unknown: viz. (8m — &)-j-3, in 
which m represents the chord of half the arc, and b the chord of the arc, an- 
swers tolerably well when the arc is not greater than a quadrant; for it ex- 
Sresses a quadrant, only 4' 20" short, and for a shorter arc the error is less; 
ut it expresses the arc 2%° short when it is equal to two quadrants. 

Example. — The chord of a circle-arc is 24 and the cosine 16. 
What is the length of the arc ? 

r = vV + (£&)*] = 20, and c > ±b ; 

and this example involves the maximum error, 
(true 1= 25.73998). Ans. 



LINES AND SUPERFICIES. 



201 



Example. — Half the chord of a circle-arc is 9, and the versed 
sine 5 ; required the length of the arc. 

(9 2 _|_ 5 2 ) _i_ 2 X 5 = 10.6 = radius of arc, 

and 10.6 — 5 = 5.6 = cosine, c<^|&; 

thenn = 180 172.27922 X 5.6 =116 o. I91 765, 

10.6 +4.5 + .0197354 



and I 



Tzr 
"180 



r= 21 - 496 =n x ~ioi+ 



957107 X 5.6 



4.5 X .0197 + 



)• 



Ans. 




To find the Area of a Sector of a Circle. 

Let A represent area of sector of salient angle (acbd a, dia- 
gram) ; I, length of its arc ; n, number of degrees in its arc ; r, radius 
of circle. 

A = 7rr 2 n-f-360 = ^r. 

But tit 2 = area of circle containing 360°. 

The area of the greater sector, or sector of 

the re-entrant angle, therefore = 

w»*(360— n) -f- 360 = ir^—^lr = tit 2 — A. 

To find the Area of a Segment of a Circle. 

Let A represent area of segment of salient angle (ab d a, dia- 
gram) ; Z, length of its arc ; n, number of degrees in its arc ; c, cosine 
of half its arc (c p, diagram) ; b, chord of arc ; and r, radius : then 

A = nr 2 n-^ 360— ±bc = ±lr— \bc. 

But the area of a circle = nr 2 . 

The area of the segment of the re-entrant angle therefore = 

7^(360 — w) -^360 -j- \bc — 7rr 2 + \bc— \lr = tu- 2 — A. 

Note. — When the arc of the segment does not contain more than 90°, 

4y /4m . , \ , v(ib 2 4- 3v 2 ) 

nearly; m being the chord of half the 



db 



. 4v /4m , , \ , 

A= ro(T +6 ) nearly ' = 

arc, and v the versed sine. And when the arc contains more than 90°, and not 
exceeding 180,° A = ~ «/ (& 2 + §^) nearly. 



202 LINES AND SUPERFICIES. 

To find the Area of a Zone. 
Rule. — Find the area of the circle containing the zone, and 
the areas of the segments on the sides of the zone, and from the 
area of the circle subtract the sum of the areas of the segments ; 
the difference will be the area sought. The rule is also applicable 
in finding the area of a zone of an ellipse. 

To find the Diameter of a Circle of which a given Zone is a part. 

d =-<A \b 2 + ( h -J jt — j yd being the diameter of the circle, 

B the greater base, and b the less, of the zone, and h the perpen- 
dicular distance between the bases. 

To find the Area of a Crescent. 

Rule. — Find the areas of the two segments formed by the 
arcs of the crescent and their chord ; subtract the less from the 
greater, and the difference will be the area of the crescent. 

To find the Side of a Square that shall contain an Area equal to 
that of a given Circle. 

Rule. — Multiply the diameter of the circle by \/ '(it -£- 4) z= 

.886228, or multiply the circumference by V^-r" ^ = -28209447 ; 
the product will be the side. 

To find the Diameter of a Circle that shall have an Area equal to 
that of a given Square. 

Rule. — Multiply the side of the square by ^(4-^-^) — 
1.128379, and the product will be the diameter required. 

To find the Diameters of Three Equal Circles the greatest that can 
be inscribed in a given Circle. 

Rule. — Multiply the diameter of the circumscribing circle by 
3 -7- f 7r 2 = .45594322 ; the product will be the diameters required. 

To find the Diameters of Four Equal Circles the greatest that can 
be inscribed in a given Circle. 

Rule. — Multiply the diameter of the given circle by 4 ~- tt 2 = 
.4052829, and the product will be the diameters sought. 



LINES AND SUPERFICIES. 203 

To find the Side of a Square inscribed in a given Circle. 

A square is said to be inscribed in a circle when all its corners 
touch the circumference. Its area is half that of a circumscribing 
square of the same circle. 

Rule. — Multiply the diameter of the circle by */-| = -| a/2i= 

.7071068, or multiply the circumference by ^/i~7-^ — .225079 ; 

the product will be the side of the inscribed square. 

Example. — A log is 28 inches in diameter at the smaller end ; 

required the side of the greatest square that can be hewn from it. 

28 X .7071068 = 19.799 inches. Ans. 

Note. — With reference to circumscribed and inscribed figures generally, 
see Centres of Surfaces, p. 291. 

To find the Diameter of a Circle that will circumscribe a given Tri- 
angle. 
Let I = longest side of triangle ; m, intermediate side ; s, shortest 
side; d, diameter of circle. 

d = - 



When two of the sides are equal, 

a being one of the equal sides, and m the unequal side. 
When all the sides are equal, 

j 2q2 2 « , ' -, 

d = . ox = —r- == 1-154 7a. 

To find the Diameter of the greatest Circle that can be inscribed in 

a given Triangle. 

l-\-m-\-s 
When two of the sides are equal, 

d=1 ™ \/(4q 2 — m 2 ) 
m~\-2a 
When all the sides are equal, 

d = \/(3a 2 ) -f- 3 = a — V 3 zz: .5 7 735a. 



204 



LINES AND SUPERFICIES. 



To divide a Circle into any number of Concentric Circles of equal areas. 
Rule. — 1. Multiply the square of the radius of the given circle 
by the number of concentric circles, less 1, required, and divide the 
product by the number of concentric circles required ; the square 
root of the quotient will be the radius of the given circle, less the 
breadth of the outermost concentric circle. 

2. Multiply the square of the radius of the given circle by the 
number of concentric circles, less 2, required, and divide the product 
by the number of concentric circles required ; the square root of 
the quotient will be the radius of the given circle, less the sum of 
the breadths of the two outermost concentric circles, &c. 

Example. — Required to divide a circle of 20 inches radius 
into two concentric circles of equal areas. 

20 2 -f- 2 = y/ 200 = 14.142 inches, radius of inner circle. 
20 — 14.142 = 5.858 inches, breadth of outer circle. 
The following diagram illustrates the principle of the foregoing 
rule, and exhibits the mechanical method of solving the problem. 
It is perhaps unnecessary to add that it is wholly immaterial into 
how many concentric circles of equal areas the given circle is to be 
divided, or whether the concentric circles are to have equal areas 
or otherwise. The radius of the given circle has only to be divided 
into the required number of aliquot parts, or proportional parts, the 
parallel lines struck, the curves drawn, and the division is accom- 
plished. 

Example. — Four men, A, B, C, and D, own a grindstone, of 
15 inches radius, equally between them. A is first to grind off 

his share, then B, then C ; 
and D is to have the re- 
mainder. Required the 
number of inches that may 
be ground from the radius 
by A, B, and C respective- 
ly, and the radius of the 
stone that will be left for D. 
^/(15«X 3 -r-4) =12.99; 
and 15 — 12.99 =: 2.01 in. 
—A. 

, V /(15 2 X2-^4)=10.60G, 
and 12.99 — 10.606 = 2.384 
inches, — B. 
v /(15 2 -^-4) = 7.5, and 
10.606 — 7.5 — 3.106 in.,— C. 

15 — (2.01 -f- 2.384 -L. 3.106) = 75 inches,— D. 

Note. — The square root of the quotient obtained by dividing the square of 
the radius of the given circle by the number of concentric circles required is 
equal to the radius of the innermost concentric circle. 




LINES AND SUPERFICIES. 



205 



To find the area of the space contained between two Concentric 
Circles. 
Rule. — Multiply the sum of the two di- 
ameters (that of the outer circle and 
that of the inner) by their difference, 
multiplied by .7854, and the product will be 
the area of the ring, or space, referred to. 

Example. — The diameter of the larger 
circle, A B, is 36 inches, and the diameter 
of the less circle, C D, is 30 inches ; required 
-the area of the ring, or space, between the 
two circles. 

36 -j- 30 = 66, and 36 — 30 =: 6 ; then 
66 X 6 X .7854=311 square inches. Ans. 




ELLIPSE. 

An ellipse is a single closed curve and an isometrical projec- 
tion of a circle. 

It is formed when a plane cuts through both sides of a cone or a 
cylinder obliquely to the base. 

It has two foci : they are on the transverse diameter, one on 
each side of the centre of the figure ; and they are equally distant 
from that point. 

An ellipse is divisible by either of its right diameters into equal 
and similar semi-ellipses, and by both into four equal and similar 
elliptic curves. 

The extremities of the two diameters and two parameters of an 
ellipse are eight points in the circumference. The summit of every 
ordinate is a point in the curve. The summits of every double 
ordinate are two points in the circumference. 

Either semi-diameter of an ellipse may be an ordinate, though 
the axis of abscissae proper is the major axis. 



Let t = transverse diameter. 
c=. conjugate diameter. 
p z=z parameter. 

fz=z distance from vertex to focus. 
gz=i space between the foci. 
e z=z eccentricity. 



Let <p — axis of abscissae. 
= axis of ordinates. 
v = greater 
x = lesser 
z = abscissa from centre. 
y = ordinate. 






; also 



18 



206 



LINES AND SUPERFICIES. 



When the transverse axis is the axis of abscissas, 
?/ 2 -f- (1 — e 2 ) = vx = tv — v 2 =ztx — x 2 = \t 2 — z 2 ; also 

When the conjugate axis is the axis of abscissaz, 

y 2 (l — e 2 ) — vx —z cv — v 2 = ex — x 2 = |c 2 — z 2 ; also 
/ 2 «rr / 2 t 2 / 2 

0y 0# 



.'.*= 



i» + vW - y 2 ) ~ ¥ - V' do 2 - y 2 ) 

to _ to — to _ 



to 



2 

a?, 



T 



, = |+r v / ( ^_^ )= ^ == ^ + , =; 



-I-^(^ 



» 



~" 2 i> 



2 = 0' — v, 



2!= r v /(i^_ 2/2) = ^ ( ^_ x=v _ ^ = ^_ ^ 



y = -V/(t;a:)=-(v/^-y 2 ) = -v/(^- a ; 2 ) = -(i v //-z 2 ). 

Example. — The eccentricity of an ellipse is 0.8, an ordinate 
is 12, and its abscissa, reckoned from the centre, is 9.6 ; required 
the axis of abscissa, and axis of ordinate . 

2y/(12 2 X (1 — -8 2 ) + 9.6 2 ) = 24, axis of abscissa?. Ans. 
12 2 X(1 — .8 2 )-f9.6 2 



4 



= 40, axis of ordinate. Ans. 



Example. — In an ellipse, a c b d, the 
transverse axis, a &, is 40, the conjugate axis, 
cd, 24, the greater abscissa, ap, 26, and the 
lesser abscissa, bp, 14 ; required the length a) 
of the ordinate ep. 




LINES AND SUPERFICIES. 207 

e 2 = 1 — (24 3 -^ 40 2 ) = 0.64, and 
V^[(l — .64) X 26X14], or \ 

|24 8 X26X14_24v/(26 X 14) 1 = 11.4473. Ans. 
\ 40 2 ~~ 40 j 

To find the Area of an Ellipse. 

A = tt£c~4 = 0.7854 t c. 
Example. — The axes of an ellipse are 40 feet and 20 feet. 
What is the area ? 

40 X 20 X .7854 = 628.32 square feet. Ans. 

To find the Length of the Circumference of an Ellipse. 
l 2 .£ 2 _ l 2 .3.e 4 l 2 .3 2 .5.g 6 l 2 . 3 2 . 5 2 . 7. € 8 
~ ^ 2 2 2 2 .4 2 2 2 .4 2 .6 2 ~~ 2 2 . 4 2 .6 2 .8 2 

l 2 .3 2 .5 2 .7 2 .9.e 10 
2 2 .4 2 .6 2 .8 2 . 10 2 •■•"*' 

But the following empirical formula, which I have been at some 
pains to devise, expresses the perimeter of an ellipse sufficiently- 
near for practical purposes in all ordinary cases of eccentricity ; 
viz. — 



«-«JC?+^)- 



And it may be proper to state that the expression furnishes I too 
short by 1-238 for ellipses of greatest and least possible eccen- 
tricities ; too short by 1-351 when c=f£; affords almost strict 
accuracy when c is not greater than |i, nor less than ±t ; and 
gives I too long by 1-274 when c = \t. It may also be proper 
to state that the empirical formula of the hand-books; viz., 

I = 7r I — -— , and which, by the way, I have seldom met with 



W 



given other than as affording strict accuracy, affords very incorrect 
results for ellipses of much eccentricity. It is correct only for the 
circle, or when t and c are equal ; and therefore gives I in excess 
in all cases. 

To find the Area of an Elliptic Segment. 

The area of an elliptic segment is to that of a corresponding 
circular segment as the axis of ordinate is to the axis of abscissa, 
nearly. 



208 LINES AND SUPERFICIES. 

Let <f> = axis of abscissa. 

= axis of ordinate. 

z = abscissa from centre. 

x == abscissa from vertex. 
B = base of corresponding circular segment. 
L — length of arc of lesser corresponding circular segment. 
A = area of lesser elliptic segment. 

B = y/(> 2 — 4z 2 ) = 2\/(<j>z — z 2 ) ; z = %<f> — x. 

When z is equal to or greater than ±B, 

T 1.5023tf>B , 3.0046fB 

Z> — — very nearly, = — — - very nearly. 

When z is less than %B, 

L= i * _ very nearly, = i-Z ± -g U. 1 

1 + - 1+— 

2<j> 2(j> 

very nearly. 
A = J (\<pL — $Bz) = 4(j) * 5 and 

— — A = area of greater elliptic segment. 

Example. — The base o of an elliptic segment is 32; the height, 
5 ; and the axis parallel to the base, 40. What is the area of the 
segment ? 

40X5 Ox 

? — 20 — v/(20 2 — 16 2 ) — — 10 — iy/(0 2 — 6 2 ) ' 

z = (25 — 2 X 5) -r 2 = 7 - 5 = it —a, 

5= ^/(25 2 — 4 X 7.5) = 20 = \/((j> — 42), 

j — 1 - 5708 X 25 + -7854 X 20 — 3.0046 X 7.5 = g3 m 



1.4 



A = 40(25X 23.174 -2X20X 7.5) = 

4X25 xxx.i*. ^/**. 

Or, in this case (see note p. 201), 



LINES AND SUPERFICIES. 209 



PARABOLA. 

A Parabola is a single open curve of two branches. It has 
but one focus ; and its eccentricity is constant, and equal to unity, 
or 1. 

It is formed, when a plane intersects a conic surface, in the di- 
rection of the base, parallel to an element of the surface, or par- 
allel to one of the sides. 

It is equated upon the principle of opposite parabolas having 
their vertices towards one another, and has both a transverse and 
a conjugate diameter ; but these lie outside the figure. The trans- 
verse is the distance between the origins of the opposite parabolas ; 
and the conjugate is tangent to the vertex of the true or contem- 
plated parabola, and is bisected by the transverse. The middle, 
of the transverse is the centre, and its projection is the axis of the 
figure. 

But, in the parabola, different abscissas on its axis, reckoned 
from the origin, are to one another as the squares of their ordi- 
nates. The parabola, therefore, may be equated independent of 
the supposed diameters and abscissas from the centre. Formulas 
will be given covering both methods. 

In geometry, any double ordinate of a parabola is the base of 
a parabola, and its abscissa, reckoned from the origin, is the axis or 
height ; thus a parabola is a plane figure having an area, a length 
of curve, &c. 

Let t = transverse diameter. 



c = conjugate diameter. 
p = parameter. 
/=. distance from origin to 
focus. 



Let x = abscissa from origin. 
z = abscissa from centre. 
y = ordinate. 



f:(lp) 2 ::x:y 2 - 



2 2 2 n ' 

* = ^=A.= — =2z * 






2 2 2 

_f lL__ ln __y 

J lt—±x~*P—~±z—2t 
18* 



210 LINES AND SUPERFICIES. 

«= - 2 -^ + ^-2+ F - ^ 2 

y=.tfpz) = 2)/(fa)= K X = J(^— 4c 2 ) = 
V/( P , -|c 2 ) =1 P(2 " 2 ~° = y/(^ - 2/0. 

Tb ./me? ^e J.rea of a Parabola. 
A = Ayx -^- 3 = 2&A -f- 3 ; 5 being the base and h the height. 



To find the Area of a Zone of a Parabola. 



A — „ .,. ; /' 



3(6 Z 

b being the greater base, d the lesser, and ^ . 
a the distance between them. 



\ 



c 



jTo jl?m? the Altitude of a Parabola, b, d, and a being given. 
h = b 2 a + (b 2 — d 2 )- 



To find the Length of the Curve of a Semi-parabola, or the Length 
of a Semi-parabola. 



1 = V O 2 + y 2 -h \ X V) > nearly. 



Note. — The common expression for the approximate length of a semi- 
parabola, viz. ^(lja; 2 -f-2/ 2 ), considerably exceeds the true length, unless y is 
large, compared with x. When x is to y as 3 to 5, it agrees with the foregoing ; 
and, when y is not less than 4 times greater than x, it is entitled to the prefer- 
ence. 



LINES AND SUPERFICIES. 211 



HYPERBOLA. 



A Hyperbola, like a parabola, is a single open curve of two 
branches, and has but one focus ; but its eccentricity is greater than 
unity, and hyperbolas may have different eccentricities. 

It is formed, when a plane intersects a conic surface in the direc- 
tion of the base, at any angle to the base between that which 
would generate a triangle and that which would generate a para- 
bola ; or, in other words, at any angle short of 90° to the base, and 
greater than that of the side to the base, — at any angle to the 
axis less than that of the side and axis. 

It is equated upon the principle of opposite hyperbolas having 
their vertices towards one another, as explained for the parabola. 

In geometry, any double ordinate of a hyperbola is the base of 
a hyperbola, and the prolongation of the transverse diameter is the 
axis, or altitude ; thus the hyperbola is a plane figure, having an 
area, a length of curve, &c. 

Let x = abscissa from origin. 



Let t = transverse diameter. 
c=z conjugate diameter. 
p = parameter. 
/= distance from origin to 
focus. 



z=z abscissa from centre. 
y z= ordinate. 
e i= eccentricity. 



,= 1+ - =1+ fl= 1+ |=( 1+ ^= 1 + 



tx + x* 
_ c 2 _ cx[c -f y/(c 2 -J- 4ff 2 )] 



- 1 p V 

= 2(2 — X) — 



f— te — t == w + *) l 

,_ % M | W \_* = W(c 2 + ±tf) t = cy/Cc 2 + 4f) 
2 V \ "*" e 2 — 1/ 2 2c 2 2p 

2c 
— — — z—U. 
P 2 

.-'t/fl-4- ^ \- *V^ + ^) - | *(<P + 4y 3 ) _ l , r 
Z —2 V V^e 2 — lJ Yc -~^J 4p —¥*-* 



212 



LINES AND SUPERFICIES. 



y = (/ — V)(tx + ar 2 ) = pa: +z (e 



_cv/(te + a*)' 



\ * " 2* 

To find the Length of a Semi-hyperbola. 

I = y^z 2 + f + \xy) nearly. 
To find the Area of a Hyperbola. 

A = — ( j-=k 4" m ) near ty 5 A being the 

altitude, 5 the base, and m the ordinate at 
one-third the altitude from the vertex. 

Note. — The common expression for the approximate length of a hyperbola : 
viz., 2yX .f , . , affords very incorrect results when x is great 

compared with y; but the empirical equation in the books, viz. : — 

A*fgf[21v(to+^)+4^to)], 

for the hyperbola, agrees very nearly with that offered abore. 




CYCLOID. 

A Cycloid is an elliptic arch, whose span is equal to the cir- 
cumference of the generating circle, and rise equal to the diameter 
of that circle ; or, in other words, it is a semi-ellipse by the trans- 
verse axis, when the axes are to each other as vr to 2. 

It is generated by a point in the circumference of a circle in the 
point of rest on a plane, when the circle makes one revolution 
from that point in a straight line. 

Its eccentricity is constant, the square of which is expressed by 



7T 2 



= .594717 + 



Its foci are in the span of the arch, one on each side of the cen- 
tre, and they are equally distant from that point : they are distant 
from their respective extremities of the curve, equal to 

nd (1 — Q _ nd— y/(W— 4<T) 



LINES AND SUPERFICIES. 213 

2d 



Its parameter is a single ordinate, and is expressed by. 



7T 



Its equation is expressed by y = 2 ya; (nd x) ^ w jjich agrees 

w ^1j y __£. y#(z — x) for the ellipse, when the diameters are, one 

to the other, as re to 2. 

An Epicycloid is a curve generated by a point in the circum- 
ference of a circle, which revolves about another circle, either on 
the convexity or concavity of its circumference. 

The cycloid, geometrically considered, has a perimeter, an area, 
and a length of curve. 

Let d , as before, equal the diameter of the generating circle ; 
then I = 4.081d, and A = 7r 2 d 2 ~ 4 = 2.46741 25c? 2 . 

Note. — I have been thus particular in treating of the cycloid, because I 
differ in my view from some writers with regard to the law of the curve. It 
is generally stated by writers on the subject, and without further specifying 
the nature of the curve, that its length is equal to 4 times the diameter of the 
generating circle, and that the area is equal to 3 times the area of that circle ; 
and they do not compute either by the rules they furnish as applicable to the 
ellipse. If the cycloid is not strictly elliptical, it is clear, without comment, 
that I am ignorant of its law, and that the foregoing is inapplicable. But to 
prove that the curve is not strictly elliptical, would be to prove — would it 
not? — that it is not generated by a point in the circumference of a circle 
rotating in a straight line on a plane. 

To find the Distance of Objects at Sea, fyc. 

The curvature of the earth, at its mean radius of 3956 stat- 
ute miles, or at 45° of latitude, is [y/(3956 2 -f l 2 ) — 3956] X 
5280 = 0.66734075 feet (8.0081 inches) in a single statute mile on 
the tangent, and is as the square of the distance or space between 
two levels. For a geographical mile on the .tangent, therefore, it 
is (6086 -^ 5280) 2 X 0.66734075 = 0.886043 feet. The mean hor- 
izontal refraction on the water is about T ^ of the curvature for 
any given distance ; thus the practical or apparent curvature is 
(1 — T \) 2 X .66734075 = 0.57541 feet for a single statute mile on 
the tangent. 

Prob. 1. — What is the tangent from the place of observation to 
the sea-horizon, the elevation being 30 feet ? 

y/(30 ~ .5754) == 7.22 miles. Ans. 

Prob. 2. — The distance from New York to Sandy Hook is 18 
miles ; at what elevation above the surface of the water, at either 
place, may the surface be seen at the other ? 

18 2 X 0.5754 = 186.43 feet. Ans. 

Prob. 3. — From an eminence 180 feet above the surface of* the 



214 



LINES AND SUPERFICIES. 



water at the place of observation, the flag of a ship is seen in the 
line of the sea-horizon, and the flag is known to be 60 feet above 
the surface where it is situated ; required the distance (tangent) 
from the observer's eye to the flag. 

27.9 miles. Ans. 

Prob. 4. — From an elevation 60 feet above the surface of the 
water at the place of observation, a vessel is seen in full view, and 
a portion of her canvas, supposed to be 20 feet above the surface 
where she lies, is seen in the line of the sea-horizon ; what is the 
distance from the point of observation to the vessel ? 



J 60 I 20 

.5754 ^J.5754 ' 



32 miles. Ans. 



STEREOMETRY, OR MENSURATION OF SOLIDS. 



OF PARALLELOPIPEDS AND CUBES. 
To find the lateral surface of a prism. 
Rule. — Multiply the perimeter of the base by the 
altitude, and the product is the lateral surface. If the 
surface of the entire figure is required, add the areas of 
the ends to the lateral surface. 

Example. — The sides of a triangular prism are each 
2£ feet wide, and the length of either side is 16 feet ; 
required its lateral surface. 

2| X 3 = 6| = perimeter of base, and 6| X 16 = 108 
square feet. Ans. 

Example. — A hexagonal prism has an altitude of 
12 feet, two of its sides are 2 feet wide each, three 
are 1£ feet wide each, and the remaining side is 9 inches ; 
required the lateral surface of the prism. 
2X2 = 4 
1.5 X 3 = 4.5 
.75 X 1 = .75 = 9.25 X 12 = 111 sq. feet. Ans. 

To find the solidity of a prism. 

Rule. — Multiply the area of the base by the altitude, and the" 
product is the solidity. 

Example. — The length of a triangular prism is 12 feet, and each 
side of its base is 2£ feet ; required its solidity. 




MENSURATION OF SOLIDS. 



215 



s/ 2^§ _j- 2.5 X ^ — 2. 165+ = sine of angle, or depth of base ; or, 

2.5 X 3 X .28868* = 2.165-j- = height of triangle ; and 

2.165 X V = 2.706-f- = area of base ; or, 

2.5 2 X .433012* = 2.706325 = area of base ; and 

2.706+ X 12 = 32.475-j- cubic feet. Ans. 

To find the solidity of a right prism or cube. 
Rule. — Cube one of its edges. 

Example. — The length of a side of a right prism is 
16 feet ; required its solidity. 

16 X 16 X 16 = 4096 cubic feet. Ans. 



To find the solidity of a parallelopipedon 
Rule. — Multiply the length by the breadth, and 
that product by the height ; the last product will 
be the solidity. • 

Example. — A slab of marble is 8 feet long, 3 
feet wide, and 6 inches thick ; required its cubic 
contents. 8 X 3 X -5 = 12 feet. Ans. 





OF PYRAMIDS. 
To find the lateral surface of a regular pyramid. 
Rule. — Multiply the perimeter of the base by half the slant 
height of the figure — half the length of a side — and the product is 
the surface of the sides. The surface of the entire figure is the sur- 
face of the sides with the area of the base added thereto. 

Example. — A triangular pyramid has a slant height of 60 feet, 
and each edge of its base is 20 feet ; required its lateral surface. 
20 X 3 X -y»- = 1800 square feet. Ans. 



To find the solidity of a pyramid. 
rtuLE. — Multiply the area of the base by | the per- 
pendicular height, and the product is the solidity. 

Example. — A quadrilateral pyramid has a perpen- 
dicular height of 21 feet, and each side of its base is 8 
feet ; what are its cubic dimensions % 



8 2 X V = 448 feet. Ans. 



* See Table of co-efficients, etc., relative to Polygons. 




216 MENSURATION OF SOLIDS. 

To find the lateral surface of a Frustum of a Pyramid, Frustum of 
a Cone, Prismoid, or Wedge. 

Rule. — Multiply the sum of the perimeters of the bases by half 
the slant height. 

Example. — Each edge of one of the bases of the frustum of a 
hexagonal pyramid is 4 feet ; each edge of the other base, 2 feet ; 
and the slant height of the frustum is 24 feet : required its lateral 

surface. 




4 X 6 + 2 X 6 X *f — 432 square feet. Ans. 

To find the solidity of the frustum of a pyramid. 
Rule. — To the square root of the product of the areas 
of the bases, add the areas of the bases, and multiply 
the sum by \ the perpendicular height of the frustum ;* 
the product will be the solidity. 

Example. — The greater base of the frustum of a 
quadrilateral pyramid is 3 feet on each side, and the less 
base is 2 feet on each side, the perpendicular height of 
the frustum is 15 feet ; what are its solid contents? 

3X3 = 9 feet = area of greater base. 

2X2 = 4 feet = area of less base. 

15 -f- 3 = 5 = £ height. Then, 

9X4 = V36 = 6 + 9 + 4= 19 X5 = 95 feet. Ans. 

OF PRISMOIDS AND THE WEDGE. 
. To find the solidity of a prismoid. 

The Rule for finding the solidity of the frustum of a pyramid is 
equally applicable to the prismoid. Or, 

Rule. — Add the areas of the ends, and four times the area of the 
mean between the ends, together, and multiply the sum by ^ the per- 
pendicular height ; the product will be the capacity or solidity. 

Example. — The bottom of a rectangular cistern is 8 feet by 6 
feet, the top is 4 by 3 feet, and the perpendicular depth is 12 feet ; 
required its capacity. 
8 .6 

4 .3 

12 . 9 -:- 2 = 6 X 4.5 = 27 X 4 = 108 = 4 area of mean ; and 
6 X 8 + 4 X 3 = 60 + 108 = 168 X -^ = 336 cubic feet. Ans. 

*For another rule, see Mensuration of Prismoids. 



MENSURATION OF SOLIDS 



217 



To find the solidity of a wedge. 
Rule. — Multiply the sum of the length of 
the edge and twice the length of the base, by 
the breadth of the base multiplied by the per- 
pendicular depth or length of the wedge, and 
divide the product by 6 ; the quotient will be the 
solidity. 



Example. — The length of the base of a 
wedge, a c, is 10 inches, the length of the edge, 
ef, is 8 inches, the breadth of the base, a b, is 
4 inches, and the perpendicular depth, d e, is 12 
inches ; required the contents. 




8-f-10X2 = 28X4X 12= 1344 -j- 6 = 224 cubic in. Ans. 



OF CYLINDERS. 

To find the convex surface of a cylinder. 
Rule. — Multiply the circumference by the length, 
and the product will be the convex surface. If the sur- 
face of the entire cylinder is required, add the areas of 
the ends to the convex surface. 

To find the solidity of a cylinder. 

Rule. — Multiply the area of an end by the length of 
the cylinder, and the product is the solidity. 

Example. — The diameter of a cylinder is 6 feet, 
and its length is 8 feet ; required its solidity or capac- 
ity. 

6 X 6 = 36 X -7854 = 28.2744 = area of end, and 
28.2744 X 8 = 226^— cubic feet. Ans. 

To find the length of a helix, or spiral, wound round a cylinder. 
Rule. — Multiply the circumference of the cylinder by the number 
of revolutions the spiral makes around it, square the product, and 
thereto add the square of the length of the cylinder ; the square root 
of the sum is the length of the spiral. The rule is applicable in 
finding the length of the thread of a screw, hand-rail to a winding 
staircase, &c, &c. 
19 




218 MENSURATION OF SOLIDS. 

OF CONES. 
To find the convex surface of a cone. 
Rule. — Multiply the circumference of the base by half the slant 
height, and the product is the surface required. If the surface of the 
entire figure is required, add the area of the base to the convex sur- 
face. 

Example. — The diameter of the base of a right cone 
is 8 feet, and the slant height is 18 feet ; required the 
convex surface. 

8 X 3.1416 = 25.1328 = circumference of base, and 
25.1328 X - 1 / = 226^ sq. feet. Ans. 

To find the solidity of a cone. 
Rule. — Multiply the area of the base by £ the per- 
pendicular height, and the product is the solidity. 

Example. — The diameter of the base of a right cone 
is 8 feet, and the perpendicular height of the cone is 15 
feet ; required its solidity. 

8 2 X -7854 X -V 5 - = 251J cubic feet. Ans. 

To find the solidity or capacity of the frustum of a cone. 

Rule. — 1. To the square root of the product of the areas of the 
ends add the areas of the ends, and multiply the sum by \ the perpen- 
dicular height of the frustum ; the product will be the solidity. 
. Or, — 2. Divide the difference of the cubes of the diameters of 
the ends by the difference of the diameters of the ends, and multiply 
the quotient by \ the perpendicular height of the frustum; this prod- 
uct multiplied by .7854 gives the solidity. 

Or, — 3. To the product of the diameters of the ends add I the 
square of the difference of the diameters, multiply the sum by .7854, 
and the product will be the mean area between the ends, which, mul- 
tiplied by the perpendicular height of the frustum, gives the solidity. 

Example. — The diameter of the larger end of a stick of round 
timber is 30 inches, that of the smaller end is 21 inches, and the 
length of the stick is 36 feet ; required its contents. 

30 inches = 2£ feet, and 21 inches = 1| feet. 




2.5 3 * 1.75 3 -r-2.5 * 1.75= 13.6875 X -\ 6 - X .7854 
129 cubic feet. Ans. 



By Rule 2 A 

n r> i o 5 2.5 X 1.75 — 4.375 + .1875 == 4.5625 X .7854 X 
By Rule 3.^ 36 = m% ^ 



MENSURATION OF SOLIDS. 



219 



Example. — The interior diameter of the larger end 
of a circular cistern is 12 feet, that of the smaller end 
is 8 feet, and the perpendicular depth of the cistern is 
14 feet ; required its capacity in cuhic feet. 
12 - 8 = 4 X 4 = 16 -r 3 = 5.333 = £ sq. of differ- 
ence of ends ; and 

12 X 8 + 5.333= 101.333 X .7854 X 14= 1114.217. 

Arts. 

Note. — For an Example under Rule 1, see Mensuration of Pyramids. All rules which 
are applicable to the measuring of a cone, or frustum thereof, are also applicable to the 
pyramid, or its frustum ; but, inasmuch as the areas of the ends of the two figures are 
not found with equal readiness, there tvill usually be a choice in the employment of rules. 




OF SPHERES. 

To find the surf dee of a sphere or 

Rule. — Multiply the diameter by the circumference ; or, multiply 
the square of the diameter by 3.1416, and the product will be the 
surface. 

To find the solidity of a sphere or globe. 

Rule. — Multiply the superficies by ^ of 
the diameter ; or, multiply the square of the 
diameter by ^ of the circumference ; or, mul- 
tiply the cube of the diameter by .5236, and 
the product is the solidity. 

Example. — Required the solidity of a 
cannon ball whose diameter is 9 inches. 

9 X 9 X 3.1416 = 254.46-f- sq. in., surface, or superficies, and 
254.46 X f = 381.7 cubic inches. Ans. 
Or, 9 X 9 X 9 = 729 X -5236 = 381.7. Ans. 




To find the convex surface of a spherical segment or zone. 

Rule. — Multiply the height by the circumference of the sphere. 

Or, Q = 2nrv = tv (Ad 2 -{- v 2 ), for the segment ; r being the radius 
of the sphere, d the diameter of the base of the segment, and v 
the height ; and 

p = 27rrA = l7r(D 2 — d 2 -{-4A 2 -f- 8hv), for the zone; D and d 
being the diameters of the bases, h the height of the zone, and v 
the height of a spherical segment whose base is equal to the less 
base of the zone. 



220 MENSURATION OF SOLIDS. 



To find the solidity of a spherical segment. 

Rule. — 1. To the square of the height of the segment, add three 
times the square of half the base, and multi- 
ply the sum by the height, multiplied by ^^^H^^s*. 
.5236, and the product is the solidity. /^^^Bl^^^v 

Or, — 2. From three times the axis of the a ,, , , p , ' M > , ll§k^ 

sphere, e f, subtract twice the height of the ill^ 

segment, and multiply the difference by the Jll fflftf i; l!|l|ilIlII 

square of the height, multiplied by .5236, lHB|.' t ,, ,,,. i yjj W^^Bf 

and the product will be the contents. jlplf, ; _li^H[ 

Note. — | a d'l -\- b cl -J- 6 c — ef, and e/X b c ^Sl^ %\j ' wjw 

Example. — The base, a d, of the segment a d c, is 18 inches, and 
the altitude, b c, is 8 inches ; required the solidity of the segment. 
8 2 -f 9 2 X 3 = 307 X 8 X -5236 == 1285.96 cubic inches. Ans 

To find the solidity of a spherical zone. 

Rule. — Add the square of the radius of one base to the square of 
the radius of the other, and thereto add J the square of the height ; 
multiply their sum by the height, multiplied by 1.5708, and the prod- 
uct will be the contents. 

Example. — The base, g h, of the zone g h d a, is 4 feet, the base, 
a d, is 3 feet, and the height, i b, is 2^ feet ; the contents of the zone 
are required. 

4 -T- 2 = 2, radius of base g h. 2T5 2 _ 6 25 _^_ 3 = 2 .0833 = 
3-j- 2 = 1.5 " ad. ^ square of height. Then, 

£2 _|_ I752 _|_ 2.083 = 8.333 X 2.5 X 1.5708 = 32.72-f cub. feet. 

Ans. 

To find the side of the greatest cube that can be cut from a given sphere. 
Rule. — Divide the square of the diameter of the sphere by 3, and 
the square root of the quotient is the side ; or, multiply the diameter 
of the sphere by .57735, and the product is the side. 

Example. — The diameter of a globe is 15 inches; required the 
eide of the greatest cube that may be cut rrom the globe. 
15 X 15 = 225 -J- 3 = V75 = 8.66 inches, or 
15 X .57735 = 8.66 inches. Ans. 



MENSURATION OF SOLIDS. 



221 



OF SPHEROIDS. 
To find the solidity of a spheroid. 
. Rule. — Multiply the square of the 
revolving axis by the fixed axis multi- 
plied by .5236, and the product is the 
solidity. 

Example. — The fixed axis, a b, of 
the prolate spheroid a c b d, is 32 inches, 
and the revolving axis, c d, is 20 inches ; 
required the solid contents. 

20 2 X ?2 X .5236 = 6702.08 cubic inches = 
6702.08 ■+■ 1728 = 3.88- cubic feet. Arts. 




To find the solidity of the segment of a spheroid. 

Rule. — When the base of the segment is parallel to the shorter axis 
of the spheroid. — From three times the length of the longer axis, 
subtract twice the height of the segment, and multiply the difference 
by the square of the height, multiplied by the square of the shorter 
axis, multiplied by .5236, and divide the product by the square of the 
longer axis ; the quotient will be the solid contents of the segment. 

Rule. — When the base of the segment is parallel to the longer axis 
of the spheroid. — From three times the length of the shorter axis 
subtract twice the height of the segment, and multiply the difference 
by the square of the height multiplied by the longer axis, multiplied 
by .5236, and divide the product by the shorter axis ; the quotient 
will be the solidity. 

Example. — The longer axis, a b, of the spheroid a c b d, being 32 
inches, and the shorter, c d, 20 inches, what is the solidity of the seg- 
ment ef a, whose base, ef, is 12 inches, and height, g a, 4 inches? 

32 X 3 — 8 = 88 X 16 X 20 2 X .5236 = 294891.52 -H 1024 = 287.98 
cub. inches. Ans. 

To find the solidity of the middle frustum of a spheroid. 
Rule. — When the ends of the frustum are parallel to the revolving 
axis. — To the square of the diameter of either end add twice the 
square of the revolving axis, and multiply 
the sum by the length of the frustum multi- 
plied by .2618 ; the product will be the so- 
lidity ; or the capacity, the frustum being a 
cask, and the measures taken of the interior. 

Example. — The diameters, ef and g h, 
of the frustum a d b c, are 18 inches each, 
19* 



7 i' 
[ah 


< 





222 



MENSURATION OF SOLIDS. 



the revolving- axis, c d, is 23 inches, and the length of the frustum, 
a b, is 20 inches ; required the cubic capacity of theTrustum. 
23- X 2 + 18* = 1382 X 20 X -2618 = 7236.152 -^ 1728 = 4.187 
cubic feet. Ans. 

OF SPINDLES AND CONOIDS. 
To find the solidity of an elliptic spindle. 
Rule. — To the square of twice the diameter of the spindle, at \ 
its length, add the square of its greatest diameter, and multiply the 
sum by the length, multiplied by .1309 ; the product will be the solid- 
ity, nearly. 

Example. — The greatest diameter, 
c d, of the elliptic spindle a c b d, is a< ^; 
12 inches, the diameter at£ its length, 
efi, is 8£ inches, and the length of the 
spindle, a b, is 40 inches ; required its solidity. 
8.5 X 2 = 17, and"l7 2 +12 2 X 40 X -1309 = 2267.188 cubic, in 




Ans. 



To find the solidity of a parabolic spindle. 
Rule. — Multiply the square 
of the middle diameter by the 
length of the spindle multiplied 
by .41888, and the product is E 
the solidity. 

Example. — The diameter, 
c d, at the middle of the para- 
bolic spindle Erf F c, is 15 

inches, and the length of the spindle, E F, is 40 inches 
solidity. 

15 2 X 40 X .41888 = 3769.92 cubic inches. Ans. 




required its 



To find the solidity of the middle frustum of a parabolic spindle. 

Rule. — To eight times the square of the greatest diameter, add 
three times the square of the least diameter, and four times the prod- 
uct of the two diameters, and multiply the sum by the length of the 
frustum, multiplied by .05236, and the product is the solidity. 

Example. — The greatest diameter, c d, of the frustum a cb d,\s 
28 inches, the least diameter (that of either end) is 20 inches, and the 
length of the frustum, a b, is 40 inches ; required the solidity. 



28 2 X 8 + 20 2 X 3 + 28 X 20 X 4 X 40 X .05236 
cubic feet. Ans. 



m 



w 



MENSURATION OF SOLIDS. 



223 



To find the solidity of a paraboloid, or parabolic 
conoid. 
Rule. — Multiply the area of the base by- 
half the altitude ; or, multiply the square of 
the diameter of the base by the altitude mul- 
tiplied by .3927, and the product will be the 
solidity. 

Example. — The diameter of the base, c d, 
of the paraboloid, c d g, is 40 inches, and its 
height, f g, is 40 inches ; required its solidity. 

40 2 X -7854 X 20 = 25132.8 cubic inches. Ans. 




To find the solidity of a frustum of a paraboloid. 
Rule. — Add the squares of the diameters of the two bases 
together, and multiply the sum by the distance between the bases, 
multiplied by .3927 ; the product will be the solidity. 

Example. — The diameter of the base, c d, of the frustum, c d b a, 
is 40 inches, the diameter of the base, a b, is 22 inches, and the height 
of the frustum,/ e, is 26 inches ; required the solidity, or capacity. 

40 2 + 22 2 = 2084 X 26 X -3927 = 21278 cubic inches. Ans. 

To find the solidity of a hyperboloid or hyperbolic conoid. 
Rule. — To the square of half the <£. 

diameter of the base add the square of 
the diameter mid way between the base 
and vertex, and multiply the sum by /^ — ' 

the distance between the base and ver- 
tex, multiplied by .5236 ; the product / 
will be the solidity. 

Example. — The diameter of the ^ :0 

base, a b, of the hyperboloid, a bf is 
40 inches, the diameter midway between the base and vertex is 26 
inches, and the altitude of the figure, e f, is 24 inches ; required its 
solidity. 

20 2 + 26 2 = 1076 X 24 X .5236 = 13521.4+ cubic in. Ans. 

To find the solidity of a frustum of a hyperboloid. 
Rule. — To the square of half the diameter of one end, add the 
square of half the diameter of the other end, and the square of the 
diameter midway between the two ends, and multiply the sum by the 
length of the frustum, multiplied by .5236 ; th^ product is the so- 
lidity. 




224 MENSURATION OF SOLIDS. 

Example. — The diameter of the base, a b, of the frustum, a b d c, 
is 40 inches, the diameter of the base, c d, is 17 inches, the diameter 
at the point equally distant from either base, is 29.1 inches, and the 
length of the frustum, e r, is 18 inches ; required the solidity. 

20 2 -f 8£* + 29~7l 2 = 1319 X 18 X .5236 = 12432 cub. in. Ans. 

Note. — The solidity or capacity of a frustum of any of the conic sections may bo 
found by first finding the mean diameter, which, on being squared and multiplied by the 
length of the frustum decreased by being multiplied by .7854, gives the cubic contents. 
Instance, the last example : — 

40 — 17 = 23 X .55* = 12.65 -f 17 = 29.65 = mean diameter, and 
29J652 X 18 X -7854 a 12428 cub. in. Ans. 

To find the surface of a cylindrical ring. 
Rule. — To the inner diameter of the ^-^TZT^< 

ring add its thickness, and multiply the sum //i^^^^^feX 

by the thickness, multiplied by 9.8696 ; the //f IrA 

product is the surface. (((L 1\\ 

Example. — The inner diameter, be, of a X \ y 1 jj 

cylindrical ring is 12 inches, and its thick- %\ J/J 

ness, ab, is three inches ; required the super- ^ -^s^F 
ficies. ^^^^^^^ 

12 -f- 3 = 15 X 3 X 9.8696 = 444 sq. inches. Ans. 

To find the solidity of a cylindrical ring. 
Rule. — To the inner diameter of the ring add its thickness, and 
multiply the sum by the square of the thickness, multiplied by 2.4674 ; 
the product is the solidity. 

Example. — The inner diameter, b c, is 12 inches, and the thick- 
$, a b, is 3 inches ; the solidity is required. 

12 -f-. 3 X 3 2 X 2.4674 = 333.1 cubic inches. Ans. 

♦ For the co efficient multipliers, and rule in detail, see Gauging, Rule 6. 



MENSURATION OF SUPERFICIES. 



225 



OF THE REGULAR BODIES. 



The Regular Bodies are five in number, viz. : 

The Tetrahedron or equilateral triangle ; a solid bounded by four 
equilateral triangles. 

The Hexahedron or cube ; a solid bounded by six equal squares. 

The Octrahedron; a solid bounded by eight equilateral triangles. 

The Dodecahedron ; a solid bounded by twelve regular and equal 
pentagons. 

The Icosahedron ; a solid bounded by twenty equilateral triangles. 

V\e following Table shows the superficies and solidity of each of the 
regular bodies, the linear edge of each being 1. 



No. of sides. 


Names. 


Superficies. 


Solidities. 


4 


Tetrahedron. 


1.73205 


0.11785 


6 


Hexahedron. 


6.00000 


1.00000 


8 


Octahedron. 


3.46410 


0.47140 


12 


Dodecahedron. 


20.64573 


7.66312 


20 


Icosahedron. 


8.66025 


2.18169 



To find the superficies of any of the regular bodies, by help of the fore- 
going table. 

Rule. — Multiply the tabular number in the column of super- 
ficies by the square of the linear edge, and the product will be the 
surface. 

Example. — The linear edge of a tetrahedron is 3 feet ; required 
the superficies. 

1.732 X 3 2 = 15.588 square feet. Ans. 

^o find the solidity of any of the regular bodies, by help of the fore- 
going table. 

Rule. — Multiply the tabular number in the column of solidities 
by the cube of the linear edge, and the product will be the solidity 
or contents. 

Example. — The linear edge of an octahedron is 2^ feet ; required 
the solidity. 

.4714 X 2.5 X* 2.5 X 2.5 = 7.3656 cubic feet. Ans. 



226 PROMISCUOUS EXAMPLES IN GEOMETRY. 



PROMISCUOUS EXAMPLES IN GEOMETRY. 

1. To find the Diameter of a Circle, or the Side of a Square, whose 
Area shall bear a given ratio to the Area of a given Circle, or given 
Square. 

D = diameter of given circle, or side of given square. 
d — diameter of required circle, or side of required square. 
a — b = given ratio of area, d = \f(aD 2 — b). 

Example. — An engraver has a drawing which may be circum- 
scribed by a square whose sides are 8 inches, and he wishes to copy 
it at •§ the size : required the sides of the square that will cir- 
cumscribe the proposed copy. y/[(2 X 8 2 ) -f- 3] = 6.532-in. Ans. 

To find the Sides of a Rectangle, Rhombus, or Rhomboid, whose 
Area shall bear a given ratio to the Area of a given similar Rec- 
tangle, Rhombus, or Rhomboid. 

L = length of given figure; H— breadth or perpendicular 
height of given figure ; I — length of required figure ; h = breadth 
or perpendicular height of required figure ; a -~- b = given ratio of 
area. 

l = ^(aL 2 + b). h = \/(aH*-±-b) 

Or, L : I : : H ; h, and H ; h \ • L ; I. 
Example. — An engraver has a drawing which may be circum- 
scribed by a rectangle whose length is 8, and breadth 6 ; and he 
wishes to copy it at half the size: required the length and breadth 
of a rectangle that will circumscribe the proposed copy. 
\/[(8 X8)-|-2] = 5.65685 = length of required rectangle. 7 . 
8 x 6-r 2 X 5.65685 — 4.24264 = breadth required. j s ' 

II. A brace is to run 3 feet on the post of a building (from angle 
to extreme) and 4 feet on the plate. What must be the length of 
the brace (extreme from shoulder to shoulder), and at what an- 
gles must the shoulders of the tenons be cut ? 

y/(4 2 -J- 3 2 ) = 5 feet, length of brace. Ans. 
3 X 86.14 -^ (5 -J- |) = 36°.91, angle for head of brace. A 



ni 



90° — 36°.91 = 53°.09, angle for foot of brace. Ans. 



III. The roof of a building 30 feet wide is to have a height 
equal to one-third the width of the building $ required the length 
of the rafters, supposing the sides of the roof are to be equal, and 
the angles for their ends. 



PROMISCUOUS EXAMPLES IN GEOMETRY. 227 

Half width of building = 15 feet ; height of roof = 10 feet. 

y/(152_L. io 2 ) = 18.03 feet, length of rafters or slant height of 
roof. Ans. 

10X86.14 -|- (18.03-J-7.5) =33°. 74, pitch of roof, or angle 
for foot of rafters. Ans. 

90° — 33°. 74 = 56°.26, angle for head of rafters if they are to 
be bevelled together ; or (90° — 56°.26) X 2 = 67°.48 (direct 
with that for the foot), if one is to rest upon the top of the other, 
or if. they are to be halved together, or if an equilateral ridge-pole 
is to be used; in which latter case, the upper and lower angles of 
the pole must be 90°-|- 56°.26 — 33°.74= 112°.52 each, and the 
lateral angles 180° — 112°.52 = 67°.48 each. But if the upper 
ends of the rafters are to be squared, by which the pole will have 
two right angles, the lateral angles of the pole must be 90° each, 
the upper angle 112°.52, as before, and the lower angle 67°.48. 
Ans. 

IV. A king-post is to run 5.1 feet on the rafter, and 5 feet on 
the beam ; required the length of the post, and the angles for its 
ends, the perpendicular h (dropped from the extremity of the run 
on the rafter to the beam) beino; 2.8 feet. 

b= v/(5.1 2 —2,8 2 ) = 4.26263 feet, distance from foot of per- 
pendicular to angle opposite post. 

p = y/[2.8 2 -4- (5 — 6) 2 ] = 2.89546 feet, length of post. ' Ans. 

86.147* ~ (5.1 -f- £6) = 33°.35, angle opposite post, or pitch of 

roof; and 90° — 33°.35 -4- 86 ' 14 X (5 — 5) _ 71 o 44 a j f 

head of post, Ans. 

180° — 71°.44— 33°.35 =75°.21, angle for foot of post (in- 
verse to that for the head). Ans.' 

The foregoing is applicable to a roof of equal or unequal sides, 
and to hip-roofs generally. 

Note. — Post perpendicular to rafter, acute angle of foot of post on out- 
side. 

Post perpendicular to beam, acute angle of head of post on outside. 

Post inclining toward opposite angle, and inner vertical angle greater than 
90°, acute angle of foot of post on outside; of head of post on inside. 

Post inclining toward opposite angle, and inner vertical angle less than 90°, 
acute angle of foot of post on outside; of head, on outside. 

Post declining from opposite angle, acute angle of foot of post on inside; 
of head, on outside. 

V. Two perpendicular walls are standing on a plane 80 ft. apart ; 
one is 20 feet high, and the other 30 ; at what distance on the plane, 



228 PROMISCUOUS EXAMPLES IN GEOMETRY. 

between them, from the base of the highest must a ladder be placed, 
so that, being inclined to either, it will reach the top ; and what 
must be the length of the ladder ? 

802 _ 302 ^ 20 2 = 5900 -h (80 X 2) = 36£ feet from base of 
highest wall. Ans. 

V (36J 2 -f 30 2 ) = 47.54 feet, length of ladder. Ans. 
Proof. 80 — 36£ = 43 J feet from base of lowest wall ; and 
V (43 1 2 + 20 3 ) = 47.54 feet, length of ladder, as before. 

A pole 100 feet in length is standing on a plane ; at what height 
must it be cut oiF, so that (the butt resting on the stump) the top 
will reach the ground 80 feet from the base of the stump ? 

80 2 

100 -loo 

- = 18 feet. Ans. 

A 

1002 — 80 2 
' 100X2 = 18 feet ' Ans ' 

_ T T)h (D — d)R , Hrf 

VI. In a Cone, j^=H; D -^ h~ A = D ^ 

= d; D being the diameter of the base, H the altitude, 

d the diameter at any given altitude h above the base, and h the 

altitude above the base to any given diameter d, or H the slant 

length of the cone, and h the slant length of the frustum ; also, 

DA (S — 5)D 3 H Sd s 

~ (S=7)D 3 = H;I) -^^S X D = ^S^ D : 

D-V g 

,,(S — 5)D 3 
f^/ ^ = d ; S being the solidity of the cone, and s the solidity 

of the frustum. 

It is required to cut from a cone, by a section parallel to the base, 
a frustum containing 16 cubic feet ; the altitude (length of axis) of 
the cone is 14 feet, and the diameter of its base 4 feet ; what must be 
the altitude of the frustum ? 

,(58.6432 — 16) X 64 „ 14 . , „ 

4-^ 5 8.6432 - = - 403X T = L41feet - Ant ' 



PROMISCUOUS EXAMPLES IN GEOMETRY. 229 

tttt T n (D — d)h (D — m)H , 

Vll In a frustum of a cone, D 77 ==m; — — A, 

( P~^ = H; ("L-y + ^D; D - (D ~^ H = i; D 

D — m H — A ' A 

being the diameter of the greater base, d the diameter of the less 
base, H the altitude of the frustum, h the altitude at any given di- 
ameter m above the greater base, and m the diameter at any given 
(T)TT T)TT 

p _ — H)d 2 .2618 -f- S = _ ^ D 2 .2618 = 

D 3 H. 

■' ^ , - = solidity of cone completed from the frustum, S being the 

D 3 H.2618 D 3 H.2618 
solidity of the frustum ; and ~ , : ~ , s :: D 3 : m% 

(J) — d)s\ 
or z 3 / (D 3 tj nftio ) = m , the diameter of the less base of a frus- 
tum whose solidity is s, cut by a plane parallel to the bases, 
from the greater end of the frustum whose solidity is S; also, 

/ s 3D 2 \ D / s 3m 2 \ m 

V (,06l8- T j ~ 2 =m ' V V^26l8~ IT)- 2 = D ' aud 
s 

■,-r.o i r- , -p. — r~naT5 = ^> the altitude of the frustum, the diame- 

(D 2 -|- m 2 -f- Dm) .2618 ' 

ter of whose less base is m, and solidity s. 
If the frustum whose solidity is s is to be taken from the smaller 

end of the given frustum, then /s/D 3 ^ = w, 

' (D— m)H 
S being the solidity of the given frustum ; and H -p. , = A 1 , 

altitude of frustum whose solidity is s, taken from smaller end of 
given frustum. 

^ A2 -(^i^) 2 = H ' and ^ H2 + (^r^) 2==A ' A be " 

ing the slant length of the frustum. 

A round stick of timber has diameter of greater end 3 feet, diame- 
ter of less end 2 feet, and length of axis 12 feet ; at what distance on 
the axis, from the greater end, must I cut this stick, by a section 
parallel to the ends, in order that the frustum, cut from the greater 
end, may contain 25 cubic feet? 

(3 — 2)25 \ 12 

3 - tf (3 3 — l 2 x 2618 j X 33^ = 3-955 feet. Arts. 

The interior dimensions of a vessel in the form of a frustum of a 
cone are, depth 12 inches, bottom diameter 10 inches, top diameter 
20 



230 PROMISCUOUS EXAMPLES IN GEOMETRY. 

6 inches ; to what depth must the vessel be filled, it standing level 
upon its bottom, in order that it may contain 1£ wine gallons ? 

(10 — 6) X 231 X 1.5\ 
^(10 3 — ^ 12 X 2618 ) = 8 - 2364 = diameter at sur- 
face of contents, and 

(10 — 8.2364) X 12 



10 



= 5.2908 inches. Ans. 



VIII. The perpendicular depth (H) of a vessel in the form of a 
frustum of a cone is 7 feet, the bottom diameter (D) 5 feet, and the 
diameter of the open mouth, or top (d) 3 feet ; the vessel is turned 
on edge, and filled with water till the level surface of the water just 
touches the uppermost poiq,t in the bottom and lowermost in the top ; 
required the quantity of water in the vessel. 

D2_ ^/(J)d)d 

- — p_\j X .2618DH = content, then 

5 X 5 — 3XV(5X3 = 13.381 h- (5 — 3) = 6.69 X 7 X 5 X .2618 
= 61.3 cubic feet. Ans. 

The same vessel, and D the open mouth, other things as before, 

D /s/(J)d)—d 2 

^ZZ~d X - 2618 ^ H = contents = 28.49 cubic feet. Ans. 

IX. A vessel 12 feet high (H), and kept constantly full of water, 
has an opening in its side 4 feet (h) above the bottom, from which a 
jet is projected ; to what distance on the plane, level with the bottom, 
is the jet projected ? 

s/ (H — h) h X 2 = projection, then 

A/ (12 — 4) X 4 = 5.657 X 2 = 11.314 feet. Ans. 



TRIGONOMETRY. 231 



TRIGONOMETRY. 



Trigonometry is a branch of Geometry. It treats of the relations 
of the sides and angles of triangles to each other. It enables us, 
having any three of the three sides *and three angles of a triangle 
given, and one of them a side, to find the rest. It treats of the 
mensuration of the angles of triangles, therefore, and of the mensu- 
ration of the sides. 

Trigonometry is of two kinds, Rectilinear and Spherical. 

The former treats of right-lined triangles, and the latter of trian- 
gles formed by the intersections of three great circles upon the sur- 
face of a sphere. 

Rectilinear trigonometry is often denominated plane trigonometry , 
and is divided into two parts, rectangular and oblique-angular. 

In rectilinear trigonometry, instead of making use of an arc of a 
circle as the measure of an angle, the sine, tangent, secant, or cosine, 
cotangent, cosecant, of that arc, is used ; and the sides of plane tri- 
angles, therefore, take these names, interchangeably, according as 
one or another is made radius of the arc supposed, and according as. 
one extremity or the other of the radius is made centre. 

The circumference of every circle is divided into 360 equal parts, 
called degrees, (°) ; each degree into 60 equal parts, called minutes, 
(') ; and each minute into 60 equal parts, called seconds, (") : 
and an angle at the centre of a circle, formed by any two sides of a 
triangle, is as many degrees, minutes and seconds, as there are de- 
grees, minutes and seconds, in that portion of the circumference that 
the sides forming that angle embrace, or may be supposed to enclose. 

And because the sides of triangles are sines, tangents, &c, of the 
arcs that measure the angles, they are also said to be sines, tangents, 
&c, of the angles that are measured by those arcs. (See Geometry 
— definitions.) 

Proposition 1. — In every right-angled triangle, if the hypotenuse 
be made radius, one of the legs will be the sine of the angle whose 
vertex is made centre, and the other will be the cosine of the same 
angle, and either extremity of the hypotenuse, or the vertex of either 
acute-angle of a right-angled triangle may be made centre. Thus, 
if, in the diagram ABC, annexed, the hypotenuse A C be ruade 
radius, and A the centre, B C will be the sine of the angle at A, 
or of the angle A, and A B will be the cosine of the same angle. And 
if A C be made radius and C the centre, A B will be the sine of the 
angle C, and B C will be the cosine of the same angle. 

Either leg, therefore, of a right-angled triangle that is the sine 




232 TRIGONOMETRY. 

of one of the acute-angles, is also the cosine of the other acute- 
angle ; and the legs of a right-angled triangle, the hypotenuse being 
radius, are sines of their opposite angles. 

Proposition 2. — In every right- 
angled triangle, if one of the legs be 
made radius, the other will be the 
tangent of the acute-angle whose 
vertex is made centre, and the hy- 
potenuse will be the secant of the 
same angle, and either leg may bt 
made radius. Thus, if A B be made 
radius, A will be centre, B C will be 
the tangent of the angle A, and the 
hypotenuse will be the secant of the 
***•••..., same angle. And if B C be made 

radius, C will be centre, A B will be 
the tangent of the angle C, and A 
will be the secant of C. 

Either leg, therefore, of a right-angled triangle that is the tangent 
of one of the acute-angles, is also the cotangent of the other ; and 
the hypotenuse being secant of the acute-angle whose vertex is made 
centre, is also cosecant of the other ; one of the legs of a right-angled 
triangle being made radius, the other is the tangent of its opposite 
angle. 

As four right-angles can be formed about the same point, there- 
fore every right-angle is equal to a quadrant of the circle, or 90°. 

The three angles of any triangle are equal to two right-angles, 
or 180°. 

The two acute-angles of a right-angled triangle are equal to a 
right-angle. 

When a particular angle of a triangle is referred to, and the 
three letters at the three angles of that triangle are used to express 
it, the letter at the angle referred to is placed in the middle ; thus, 
both the angle and the triangle are indicated : in the expression, 
The angle ABC, the angle B, in the triangle A B 0, is meant. 

In the annexed table of natural sines, cosines and tangents, to given 
angles, the numbers, though not so marked, are considered as deci- 
mals to radius 1. 

The natural sine of 90°, therefore, is 1 ; and the natural cosine 
of 90° is 0. 

The tangent of 45° is 1, and so is the cotangent. 
The tangent of 90° is infinite. 

If two angles of a triangle are given, the other is said also to be 
given, for it is the difference between the sum of the two and 180°. 
If one of the acute-angles of a right-angled triangle be given, all 
the angles of that triangle are said to be given, for the sum of the 



TRIGONOMETRY. 



233 



three angles of any triangle, minus the sum of any two, is equal the 
third and the right-angle is known for 90°. 

In trigonometrical expressions and formulas, the following abbre- 
viations and contractions are often made use of, viz. : 

R, for tabular Radius, natural or logarithmic sine of 90°, or of 
the right-angle, (1.)* 

sin A, (or any other letter,) for sine of the angle A, or whatever 
other letter. 

cos A, for cosine of the angle A. 

tan A, for tangent of the angle A. 

sec A. for secant of the angle A. 

cosec A, for cosecant of the angle A. 

cot A, for cotangent of the angle A. 

coversin, for coversed sine. 

sin A comp B, for sine of the angle A, the complement of which 
angle is the angle B. That is, the angle A subtracted from 90°, the 
difference will be the angle B." 

These expressions, such as sin A, tan B, &c, or their equivalents, 
when used as terms in the statement of a problem, or its solution, 
are to be taken as referring to the natural or tabular sine, tangent, 
&c, of the angle indicated, and not to the sides taking those names. 
Sometimes the contraction nat. or tab. is prefixed, (as nat cos B,) 
but usually it is omitted. 

In any right-angled plane triangle, ABC — 

A being one of the acute angles, and C the other ; 



sin A = cos C 
tan A = cot C 
sec A = cosec C 



cos A = sin C. 
cot A = tan C. 
cosec A = sec C. 



And of the 


same 


angle, A or C, considerec 


separately 


» 


tan 


X 


cos 


= sin 


cot 


X sin 


= 


cos 


cos 


_i- 


cot 


= sin 


sin 


-j-tan 


= 


COS 


tan 


_j- 


sec 


= sin 


cot 


-r- cosec = 


COS 


R(l) 


T 


cosec = sin 


R 


-7- sec 


= 


COS 


sin 


X 


sec 


= tan 


cosec 


X cos 


= 


cot 


sin 


-f- 


cos 


= tan 


cos 


— sin 


= 


cot 


sec 


-j- 


cosec 


= tan 


cosec 


-f-sec 


= 


cot 


R 


— 


cot 


= tan 


R 


-r-tan 


= 


cot 


cosec 


Xtan 


= sec 


cot 


X sec 


:= 


cosec 


tan 


-L. 


sin 


= sec 


sec 


-r- tan 




cosec 


cosec 


-7- 


cot 


= sec 


cot 


—- cos 




cosec 


R 


-r- 


cos 


= sec 


R 


-j- sin 


= 


cosec 



* The logarithmic radius or tabular sine of 90° is 10 : the natural radius or tabular sine 
of 90° is 1. A table of logarithmic sines, tangents, &c, is called a table of artificial sines, 
tangents, &c. 2vJ * 



234 



TRIGONOMETRY. 




In any right- angled plane triangle, 
ABC, (Fig.) 

B C -7- A C (tan -J- sec) = sin A. 
B C -J- A B (sin -f- cos) = tan A. 
AC-rAB(R -J- cos) = sec A. 
A B -4- A (tan — sec) = sin 0. ■ 
AB-f BC (sin'-f- cos) = tan C. 
AC-i-BC(R -f- cos) = sec C. 

Or, if we denominate the longest 
side of the triangle hypotenuse, and 
of the other two sides make one 
base and the other perpendicular; 
then — 



perp 
hyp. 

base 

hyp- 

perp. 
base 



* = nat sin of angle opposite perp., or nat cos of angle opposite base. 

= nat sin of angle opposite base, or nat cos of angle opposite perp. 

= nat tan of angle opposite perp., or nat cot of angle opposite base. 

= nat tan of angle opposite base, or nat cot of angle opposite perp. 



r — - = nat sec of angle opposite perp., or nat cosec of angle opposite base. 

byp. 

= nat sec of angle opposite base, or nat cosec of angle opposite perp. 

With reference to the sides of right-angled triangles, considered 
in connection with the natural sines, tangents, &c, of their opposite 
and included angles : — 



Preceding Fig. 



A X sin A = B C. 
A B X tan A = B C. 
ABXsecA =A0. 
B C X cosec A = A C. 
A X cos A = A B. 
B C X cot A = A B. 



A C -7- sec A = A B. 
AC-r cosec A = B C. 
AB-r-cotA =B0. 
A B -J- cos A = A C. 
B -r- sin A = A 0. 
B C -T- tan A = A B. 



In every plane triangle, — right-angled, acute-angled or obtuse- 
angled, — the natural sines of the angles are to each other as the 
opposite sides. 

Thus, in the triangle ABC, annexed. 






TRIGONOMETRY. 



235 



sin A : sin B : : B : A C. 
sin B : A C : : sin C : A B. 
B C : A B : : sin A : sin C. 

In every plane triangle, the 
sum of the squares of the sides 
adjacent any angle, minus the 
square of the side opposite that j^ 
angle, divided by twice the 
product of the sides adjacent, equals the natural cosine of that 
angle. 

Thus, 




AB 2 -f-A C 2 — BC 2 
A B X AC X 2 

AB 2 -f- BC 2 — AC 2 
2 A~B~X B C 



= cos A. 



= cos B. 



AC 2 + BC-AB 2 
2ACX BC 



= cos C. 



Or, if the sides of the triangle be of unequal lengths, and we de- 
nominate them according to their relative lengths, as longest, medi- 
ate, shortest, we have the following more distinctive formulas : 



longest 2 -|- med 2 — short 2 

longest X med X 2 
longest 2 -f- short 2 — med 2 



= nat cos smallest angle. 
= nat cos mediate angle. 



long X short X 2 
med 2 -\- shortest 2 — longest 2 

mediate X shortest X 2 = nat cos § reatest an § le ' 
*/ (1 — sin 2 ) of any angle = cos of that angle. 
1 — (2 X sin 2 ) of £ any angle = cos of that angle. 
That is, A being the angle — 

1 — 2 sin 2 of £ A = cos A. 

In every plane triangle, as the sum of any two sides is to their 
difference, so is the natural tangent of half the sum of the angles 
opposite those sides to the natural tangent of half their difference ; 
and half the sum of the two angles, plus half their difference, equals 
the greater of the two. and half the sum of the two angles, minus 



236 



TRIGONOMETRY. 



half their difference equals the less, the greater angle being op- 
posite the longer side. 

Thus, in the oblique-angled triangle, ABC, (Fig.) 

A C + B C : A C * B : : tan 1±5 . tan ^£_ % and 

A+B A^B 

— 2 — + — 2~ ~ B ' 



B 



A + B _A 

2 



AB + AC : AB — AC :: 

tan i(C + B) :tan i(C — B),^ 
and 




i(0 + B) + 4(0 - B) = C, and i(0 + B) - 4(0 - B) = B. 
A B + B C : A B — B C : : tan £(C + A) : tan £(C — A), and 
C + A C — A - C4-A C — A 



A - C + A 

— = C, and — k — 



= A. 



SOLUTIONS, — Right-angled Triangles. 




ABC, the triangle, 
A C, hypotenuse, 

A B and B C, legs ; B, the right 
angle. 

The hypotenuse and angles given, to 
find the legs. 

Suppose A C 51 feet, A 28° 10 7 , A.< 
C (consequently) 90° — 28° 10' = 
61° 50'. 

The sines of angles are to each '**-•.... 

other as the sides opposite those - 

angles. 

Turning, therefore, to the tables of natural sines, &c, we find, in 
the column of sines, against angle 28° 10', sine .47204, consequently — 

R AC sin A BC 

1 : 51 : : .47204 : 24.1 feet. 

Having now A C and B C, side A B may be found by the rules in 
geometry ; or, turning again to angle 28° 10', in the tables, we find. 



TRIGONOMETRY. 



237 



against that angle, cosine .88158* ; and as the cosine of one of the 
acute angles of a right-angled triangle is the sine of the other acute 
angle, consequently, 

R A C sin C A B. 
1 : 51 : : .88158 : 45 feet. 

r 

The angles and one leg given, to find the hypotenuse and other leg. 

As before, C = 61° 50', and A = 
90° — 61° 50' = 28° 10' ; A B = 
45 feet. 




AB 

45 : 



sec A 
1.13433 



AC 
51. 



The hypotenuse and one leg given, to find the angles and other leg. 
Let A C = 85 and B C = 58 : then — 



AC R 

85 : 1 



BC 

58 : 



sin A 

.68235. 



Turning, now to the tables of nat. sines, &c, we find, in the col- 
umn of sines, against the number 68235, or the number nearest 
thereto, angle 43°. We therefore have — 

A C : R : : B C r sin A, 43° ; and 

90° — 43° = 47° = angle C. 

R AC sinC AB 

1 : 85 : : .73135 : 62.16. 

The legs given, to find the angles and hypotenuse. 
Let A B = 54.7, and B C = 32 : 



AB 

54.7 



BC 

32 



tan A 
.58501. 



Turning now to the tables of natural sines, tangents, &c, we find 
in the column of tangents 58501, or a number near thereto, and 



* When an angle 13 greater than 45° the sine, tangent, &c, of its complement i3 used. 



238 



TRIGONOMETRY. 



against that number we find 30° 20' ; the angle A, therefore, is 30° 
20', and the angle C is 

90° — 30° 20' = 59° 40'. 
sin A BC R AC 
.50503 : 32 : : 1 : 63.36*. 
Or, sin 90° : B C : : sec 59° 40' : AC 
1 32 1.98008 63.36. 



SOLUTIONS, — Oblique-angled triangles. 

Let A B C be the triangle : (Fig. annexed.) 

The angles and one side given, to find the -other sides. 

Suppose A 26° 40', B 56°, C 
180° — (26° 40' + 56°) = 97° 
20', and A C 40 feet. 

sin B : A C : : sin A : B C. 




The angle B is greater than 
45° ; its sine, therefore, is the 
vg cosine of its complement, or the 
cosine of what it lacks of 90°. 
Its sine, therefore, is the cosine of 90° — ■ 56° = 34°. Turning now 
to 34° in the tables, we find, against that angle, cos .82904 ; .82904, 
consequently, is the sine of 56°, and 

sin B AC sin A B C 
.82904 : 40 : : .44880 : 21.65 feet. 
sinB : AC : : sinC : AB. 
The angle C is not only greater than 45°, but it is greater than 
90°. Its sine, therefore, is the cosine of the difference of its supple- 
ment and 90°. It is the cosine of the difference of 180° — 79° 20' 
s= 82° 40' and 90° = 7° 20'. As in the preceding case, its sine, 
therefore, is the cosine of the difference between itself and 90°. 

Turning now to angle 7° 20', in the tables, we find its cosine 
.99182 : therefore we have — 



sin B AC 

.82904 : 40 



sin C A B 

.99182 : 47.85 feet. 



Two sides and an angle opposite to one of them given, to find the other 
angles, and other side. 

As before, A B 47.85, B C 21.65, C 97° 20'. 

A B : sin C : : B C : sin A, 26° 40' ; and 
180° — (97° 20' + 26° 40') = 56° = B. 



TRIGONOMETRY. 



239 



sin C : A B : : sin B : A C ; or, 
sin A : B : : sin B : A 0. 



Two sides and their contained angle given, to find the other angles 
and side. 

Suppose A B 47.85, B C 21.65, B 56°. 

A B -J- B C : A B ^ B C : : tan £(A -\- 0) : tan £(A ^ C). 

180° — B = R (1) • 

tan i (A + C) — tan 2 tftn 9QO ^ ^ lg()0 _ B) 

tan £(180° — B) = cot 90° ^ £(180° — B) ; therefore, 
AB-f BO AB^BC tan 62° tan 

69.5: 26.2 :: 1.88073 : .70899, 35° 20'; and 

MA + C) i(A^C) C 4(A + C) i(A^O) A 

62° -f- 35.20 =97° 20' and 62° — 35° 20' = 26°40'. 

sin A : B O : sin B : A C, or 
sin O : A B : : sin B : A C. 



The three sides given, to find the 
angles. 

Let A B C be the triangle. 
A B 47.85, A C 40, B C 21.65. 




AB 2 -f AC 2 - 
34204 



BC* 



A B X A C X 2 = cos A. 

3828 .89365, 26° 40'. 



B C : sin A : : A C : sin B. 

21.65 .44880 . 40 .82919 = cos comp B, = 90° — 34° = 56° = B. 



180° — 26° 40' -f- 56° = 97° 20' = C. 



Or,AB J -fBCf-ACr-r-ABXBCX2= cosB 



1158.34 



2071.9 .55907 = sin comp B 

90° — 34° = 56° = B. 



Or,£(AB-fAC + BC)XBC^£(BA+AC + BC)-i-ABXAC 

= cos 2 ^A. 

54.75 X 33.1 -v- 1914 = V- 94682 = .97305 = cos £ A, 13° 20 7 ; and 
13° 20' X 2 = 26° 40' = A. 



240 TKIGONOMETRY. 

AB+AO+BO A c ^ AB + AC + BC ± TW ^^ = 

2 2 

2 B 
cos 2« 

ab + a 2 c + BC xab/ b + a ; + b ^acxW = 

COS 2 £ C. 

In general, however, when the three sides of a triangle are given, 
and the angles are required, it is customary to drop a perpendicular 
upon the longest side of the triangle, or one that will fall within the 
figure,, whereby the triangle is divided into two right-angled tri- 
angles ; and then to find the angles, by the rules for right-angled 
triangles. In this case the sum of the two vertical angles will be 
equal the angle at the vertex. For rules for dropping the perpen- 
dicular, see Geometry — Triangles. 

To find the angles of a right-angled triangle approximately by means 
of the sides, or without the aid of trigonometrical tables. 

Let r represent the hypotenuse, s the shortest side, c the longest 
side, A the smallest angle, or angle opposite the shortest side. 

86.13961s , ^ 

A = r4 _i c4 _'- T* vep y nearl ? •* 

Example. — What is the smaller of the two acute angles of a 
right-angled triangle whose hypotenuse is 100, shortest side 48.735, 
and longest side 87.321 ? 

86.13961X48.735 29 o. u8j OT 29 o 8 , 6 8*, 



100 -j- 43.6605 -f 0.3654 
true angle = i 

90° — 29°.148 = 60°.853 or 60° 51' 7" = greater acute angle. 



true angles 29° 10'. Ans. 



* See Circle, length of arc of, &c. 



TRIGONOMETRY. 



2il 



TABLE OF NATURAL SINES, COSINES, AND TANGENTS. 



D. M. 


Sine. 


Cosine. 


Tangent. 


D. M. 


Sine. 


Cosine. 


Tangent. 


1 


00029 


10000 


00029 


7 


12187 


99255 


12278 


5 


00145 


10000 


00145 


10 


12476 


99219 


12574 


10 


00291 


10000 


00291 


20 


12764 


99182 


12869 


20 


00582 


99998 


00582 


30 


13053 


99144 


13165 


30 


00873 


99996 


00873 


40 


13341 


99106 


13461 


40 


01164 


99993 


01164 


50 


13629 


99067 


13758 


50 


01454 


99989 


01455 


8 


13917 


99027 


14054 


1 


01745 


99985 


01745 


10 


14205 


98986 


14351 


10 


02036 


99979 


02036 


20 


14493 


98944 


14648 


20 


02327 


99973 


02328 


30 


14781 


98902 


14945 


30 


02618 


99966 


02619 


40 


15069 


98858 


15243 


40 


02908 


99958 


02910 


50 


15356 


98814 


15540 


50 


03199 


99949 


03201 


9 


15643 


98769 


15838 


2 


03490 


99939 


03492 


10 


15931 


98723 


16137 


10 


03781 


99929 


03783 


20 


16218 


98676 


16435 


20 


04071 


99917 


04075 


30 


16505 


98629 


16734 


30 


04362 


99905 


04366 


40 


16792 


98580 


17033 


40 


04653 


99982 


04658 


50 


17078 


98531 


17333 


50 


04943 


99878 


04949 


10 


17365 


98481 


17633 


3 


05234 


99863 


05241 


10 


17651 


98430 


17933 


10 


05524 


99847 


05533 


20 


17937 


98378 


18233 


20 


05814 


99831 


05824 


30 


18224 


98325 


18534 


30 


06105 


99813 


06116 


40 


18509 


98272 


18835 


40 


06395 


99795 


06408 


50 


18795 


98218 


19136 


50 


06685 


99776 


06700 


11 


19081 


98163 


19438 


4 


06976 


99756 


06993 


10 


19366 


98107 


19740 


10 


07266 


99736 


07285 


20 


19652 


98050 


20042 


20 


07556 


99714 


07578 


30 


19937 


97992 


20345 


30 


07846 


99692 


07870 


40 


20222 


97934 


20648 


40 


08136 


99668 


08163 


50 


20507 


97875 


20952 


50 


08426 


99644 


08456 


12 


20791 


97815 


21256 


5 


08716 


99619 


08749 


10 


21076 


97754 


21560 


10 


09005 


99594 


09042 


20 


21360 


97692 


21864 


20 


09295 


99567 


09335 


30 


21644 


97630 


22169 


30 


09585 


99540 


09629 


40 


21928 


97566 


22475 


40 


09874 


99511 


09923 


50 


22212 


97502 


22781 


50 


10164 


99482 


10216 


13 


22495 


97437 


23087 


6 


10453 


99452 


10510 


10 


22778 


97371 


23393 


10 


10742 


99421 


10805 


20 


23062 


97304 


23700 


20 


11031 


99390 


11099 


30 


23345 


97237 


24008 


30 


11320 


99357 


11394 


40 


23627 


97169 


24316 


40 


11609 


99324 


11688 


50 


23910 


97100 


24624 


50 


11898 


99290 


11983 











21 



242 



TRIGONOMETRY. 



D. M. 


Sine. 


Cosine. 

97030 


Tangent. 


D. M. 


■ Sine. 

37191 


Cosine. 

92827 


Tangent. 


14 


24192 


24933 


21 50 


40065 


10 


24474 


96959 


25242 


22 


37461 


92718 


40403 


20 


24756 


96887 


25552 


10 


37730 


92609 


40741 


30 


25038 


96815 


25862 


20 


37999 


92499 


41081 


40 


25320 


96742 


26172 


30 


38268 


92388 


41421 


50 


25601 


96667 


26483 


40 


38537 


92276 


41763 


15 


25882 


96593 


26795 


50 


38805 


92164 


42105 


10 


26163 


96517 


27107 


23 


39073 


92050 


42447 


20 


26443 


96440 


27419 


10 


39341 


91936 


42791 


30 


26724 


96363 


27732 


20 


39608 


91822 


43136 


40 


27004 


96285 


28046 


30 


39875 


91706 


43481 


50 


27284 


96206 


28360 


40 


40141 


91590 


43828 


16 


27564 


96126 


28675 


50 


40408 


91472 


44175 


10 


27843 


96046 


28990 


24 


40674 


91355 


44523 


20 


28123 


95964 


29305 


10 


40939 


91236 


44872 


30 


28402 


95882 


29621 


20 


41204 


91116 


45222 


40 


28680 


95799 


29938 


30 


41469 


90996 


45573 


50 


28959 


95715 


30255 


40 


41734 


90875 


45924 


17 


29237 


95630 


30573 


50 


41998 


90753 


46277 


10 


29515 


95545 


30891 


25 


42262 


90631 


46631 


20 


29793 


95459 


31210 


10 


42525 


90507 


46985 


30 


30071 


95372 


31530 


20 


42788 


90383 


47341 


40 


30348 


95284 


31850 


30 


43051 


90259 


47698 


50 


30625 


95195 


32171 


40 


43313 


90133 


48055 


18 


30902 


95106 


32492 


50 


43575 


90007 


48414 


10 


31178 


95015 


32814 


26 


43837 


89879 


48773 


20 


31454 


94924 


33136 


10 


44098 


89752 


49134 


30 


31730 


94832 


33460 


20 


44359 


89623 


49495 


40 


32006 


94740 


33783 


30 


44620 


89493 


49858 


50 


32282 


94646 


34108 


40 


44880 


89363 


50222 


19 


32557 


94552 


34433 


50 


45140 


89232 


50587 


10 


32832 


94457 


34758 


27 


45399 


89101 


50953 


20 


33106 


94361 


35085 


10 


45658 


88968 


51319 


30 


33381 


94264 


35412 


20 


45917 


88835 


51688 


40 


33655 


94167 


35740 


30 


46175 


88701 


52057 


50 


33929 


94068 


36068 


40 


46433 


88566 


52427 


20 


34202 


93969 


36397 


50 


46690 


88431 


52798 


10 


34475 


93869 


36727 


28 


46947 


88295 


53171 


20 


34748 


93769 


37057 


10 


47204 


88158 


53545 


30 


35021 


93667 


37388 


20 


47460 


88020 


53920 


40 


35293 


93565 


37720 


30 


47716 


87882 


54296 


50 


35565 


93462 


38053 


40 


47971 


87743 


54673 


21 


35837 


93358 


38388 


50 


48226 


87603 


55051 


10 


36108* 


93253 


38721 


29 


48481 


87462 . 


55431 


20 


36379 


93148 


39055 


10 


48735 


87321 


55812 


30 


36650 


93042 


39391 


20 


48989 


87178 


56194 


40 


36921 1 92935 


39727 


30 


49242 


87036 


56577 



TRIGONOMETRY. 



243 



D. M. 


Sine. 


Cosine. 


Tangent. 


D. M. 


Sine. 


Cosine. 


Tangent. 


29 40 


49495 


86892 


"57962~ 


37 30 


60876 


79335 


767,33 


50 


49748 


86748 


57348 


40 


61107 


79158 


77196 


30 


50000 


86603 


57735 


50 


61337 


78980 


77661 


10 


50252 


86457 


58124 


38 


61566 


78801 


78129 


20 


50503 


86310 


58513 


10 


61795 


78622 


78598 


30 


50754 


86163 


58904 


20 


62024 


78442 


79070 


40 


51004 


86015 


59297 


30 


62251 


78261 


79544 


50 


51254 


85866 


59691 


40 


62479 


78079 


80020 


31 


51504 


85717 


60086 


50 


62706 


77897 


80498 


10 


51753 


85567 


60483 


39 


62932 


77715 


80978 


20 


52002 


85416 


60881 


10 


63158 


77531 


81461 


30 


52250 


85264 


61280 


20 


63383 


77347 


81946 


40 


52498 


85112 


61681 


30 


63608 


77162 


82434 


50 


52745 


84959 


62083 


40 


63832 


76977 


82923 


32 


52992 


84805 


62487 


50 


64056 


76791 


83415 


10 


53238 


84650 


62892 


40 


64279 


76604 


83910 


20 


53484 


84495 


63299 


10 


64501 


76417 


84407 


30 


53730 


84339 


63707 


20 


64723 


76229 


84906 


40 


53975 


84182 


64117 


30 


64945 


76041 


85408 


50 


54220 


84025 


64528 


40 


65166 


75851 


85912 


33 


54464 


83867 


64941 


50 


65386 


75661 


86419 


10 


54708 


83708 


65355 


41 


65606 


75471 


86929 


20 


54951 


83549 


65771 


10 


65825 


75280 


87441 


30 


55194 


83389 


66189 


20 


66044 


75088 


87955 


40 


55436 


83228 


66608 


30 


66262 


74896 


88473 


60 


55678 


83066 


67028 


40 


66480 


74703 


88992 


34 


55919 


82904 


67451 


50 


66697 


74509 


89515 


10 


56160 


82741 


67875 


42 


66913 


74314 


90040 


20 


56401 


82577 


68301 


10 


67129 


74120 


90569 


30 


56641 


82413 


68728 


20 


67344 


73924 


91099 


40 


56880 


82248 


69157 


30 


67559 


73728 


91633 


50 


57119 


82082 


69588 


40 


67773 


73581 


92170 


35 


57358 


81915 


70021 


50 


67987 


73333 


92709 


10 


57596 


81748 


70455 


43 


68200 


73135 


93252 


20 


57833 


81580 


70891 


10 


68412 


72937 


93797 


30 


58070 


81412 


71329 


20 


68624 


72737 


94345 


40 


58307 


81242 


71769 


30 


68835 


72537 


94896 


50 


58543 


81072 


72211 


40 


69046 


72337 


95451 


36 


58779 


80902 


72654 


50 


69256 


72136 


96008 


10 


59014 


80730 


73100 


44 


69466 


71934 


96569 


20 


59248 


80558 


73547 


10 


69675 


71732 


97133 


30 


59482 


80386 


73996 


20 


69883 


71529 


97700 


40 


59716 


80212 


74447 


30 


70091 


71325 


98270 


50 


59949 


80038 


74900 


40 


70298 


71121 


98843 


37 


60182 


79864 


75355 


50 


70505 


70916 


99420 


10 


60414 


79688 


75812 


45 


70711 


70711 


1. 


20 


60645 


79512 


76272 











244 



TRIGONOMETRY. 



Note. — The foregoing Tables present sines, cosines, and tangents, calculated for every 
degree anl ten minutes of the quadrant. To furnish tables calculated for a smaller divis- 
ion of.the circle than ten minutes, as for five minutes, or for one minute, would occupy too 
much space in this work. Besides, calculations to the sixth of a degree, for many practi- 
cal purposes, are sufficiently minute. If, however, greater precision is desired, the tables 
furnish a very simple means whereby to obtain it, and to almost any extent short of 
scientific minuteness and accuracy. 

Suppose, for instance, in the solution of a problem, the sine .64380 appears : now, on 
turning to the tables, the nearest sine we find to this is .64279, the sine of 40° ; while the 
next nearest is .64501, the sine of 40° 10' ; now, .64279 -f .64501 = 1.28780 -f- 2 = .64390 
= sin 40° 5' 5 ,64390, therefore, is the sine of an angle a trifle less than 40° 5', but nearer 
to that angle than to any other of full minutes. 

Again, suppose the sine .51433 appears ; the nearest sine in the tables to this is .51504, 
which is the sine of 31° ; and the next nearest is .51254, the sine of 30° 50' ; .51504 -f 
.51254 = 1.02758 -4- 2 = .51379 = sin 30° 55', and .51504 + .51379 = 1.02883 -p 2 == 
.51442 = sin 30° 57£' ; .51433, therefore, is the sine of 30° 57', very nearly. 

The foregoing principles are also applicable to the cosines and tangents, and in the 
same manner. 

The versed sine of any angle = 1 — cosine of that angle. 

The coversed sine of any angle = 1 — sine of that angle. 

The chord of any angle = sine of £ of that angle X 2. 
Example. = The sides of a right-angled triangle are 3, 4 and 5 
feet ; required a side of the greatest square that may be cut from the 
triangle. • 

A right line that bisects the right angle and extends to the hypot- 
enuse is equal to and becomes the diagonal of the required square, 

and the diagonal of any square multiplied by —y^ = a side of that 

square; then 5 : 1 :: 3 : .6 = sin 36° 51',. and 

Sin 90° + 36 c 51' — 45° = .9899 : 3 ; : sin 90° — 36° 51' = .8 : 2.4245 
= diagonal, and 2.4245 X .70711 = 1.7144= side of square. Ans. 

Or sin 90° — 36° 5r(.8) : 2.4245 : sin 45°(.70711) : 2.143, and 
(5 — 2.143 2 — 0245 2 ) -^ 4 X 2 = .2855 + t 2 = 2.2855, and 



^(5 — 2.143 — 2.2855 ) = 1.714 feet, side of square. Ans. 

Recapitulation. — When the given angle is greater than 45°, 
its sine is expressed by the cosine of the angle which is the differ- 
ence of 90°and the given angle : thus the sine of 46° is the tabular 
cosine of 90— 46 2= cosine of 44° ; and the sine of 62° 50' is the cosine 
of 90 — 62° 50' = tabular cosine of 27° 10', &c. 

To obtain the tangent of an angle that is greater than 45°, divide 
the tabular cosine of the angle which expresses the difference of 90° 
and the given angle by the tabular sine of that difference ; thus the 

„- nn cos (90— 46)= cos 44° 71934 

tangent of 46° = ^ . ' = eE7Tg = 1 - 0305 ' and 

sin 44° - 6946b 

the tangent of 

00 * — sin 31° 20' 52002 ' 



TABLE OE SQUARES, CUBES, SQUARE AND CUBE ROOTS. 245 



Number. 


Square. 


Cube. 


Square Koot. 


Cube Root. 


1 


1 


1 


1.0000000 


1.0000000 


2 


4 


8 


■ 1.4142136 


1.2599210 


3 


9 


27 


1.7320508 


1.4422496 


4 


16 


64 


2.0000000 


1.5874011 


5 


25 


125 


2.2360680 


1.7099759 


6 


36 


216 


2.4494897 


1.8171206 


7 


49 


343 


2.6457513 


1.9129312 


8 


64 


512 


2.8284271 


2.0000000 


9 


81 


729 


3.0000000 


2.0800837 


10 


100 


1000 


3.1622777 


2.1544347 


11 


121 


1331 


3.3166248 


2.2239801 


12 


144 


1728 


3.4641016 


2.2894286 


13 


169 


2197 


3.6055513 


2.3513347 


14 


196 


2744 


3.7416574 


2.4101422 


15 


225 


3375 


3.8729833 


2.4662121 


16 


256 


4096 


4.0000000 


2.5198421 


17 


289 


4913 


4.1231056 


2.5712816 


18 


324 


5832 


4.2426407 


2.6207414 


19 


361 


6859 


4.3588989 


2.6684016 


20 


400 


8000 


4.4721360 


2.7144177 


21 


441 


9261 


4.5825757 


2.7589243 


22 


484 


10648 


4.6904158 


2.8020393 


23 


529 


12167 


4.7958315 


2.8438670- 


24 


576 


13824 


4.8989795 


2.8844991 


25 


625 


15625 


5.0000000 


2.9240177 


26 


676 


17576 


5.0990195 


2.9624960 


27 


729 


19683 


5.1961524 


3.0000000 


28 


784 


21952 


5.2915026 


3.03G5889 


29 


841 


24389 


5.3851648 


3.0723168 


30 


900 


27000 


5.4772256 


3.1072325 


31 


961 


29791 


5.5677644 


3.1413806 


32 


1024 


32768 


5.6568542 


3.1748021 


33 


1089 


35937 


5.7445626 


3.2075343 


34 


1156 


39304 


5.8309519 


3.2396118 


35 


1225 


42875 


5.9160798 


3.2710663 


36 


1296 


46656 


6.0000000 


3.3019272 


37 


1369 


50653 


6.0827625 


3.3322218 


38 


1444 


54872 


6.1644140 


3.3619754 


39 


1521 


59319 


6.2449980 


3.3912114 


40 


1600 


64000 


6.3245553 


3.4199519 


41 


1681 


68921 


6.4031242 


3.4482172 


42 


1764 


74088 


6.4807407 


3.4760266 



21 



246 TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 



Number. 


Square. 


Cube. 


Square Root. 


Cube Hoot. 


43 


1849 


79507 


6.5574385 


3.5033981 


44 


1936 


85184 


6.6332496 


3.5303483 


45 


2025 


91125 


6.7082039 


3.5568933 


46 


2116 


97336 


6.7823300 


3.5830479 


47 


2209 


103823 


6.8556546 


3.6088261 


48 


2304 


110592 


6.9282032 


3.6342411 


49 


2401 


117649 


7.0000000 


3.6593057 


50 


2500 


125000 


7.0710678 


3.6840314 


51 


2601 


132651 


7.1414284 


3.7084298 


52 


2704 


140608 


7.2111026 


3.7325111 


53 


2809 


148877 


7.2801099 


3.7562858 


54 


2916 


157464 


7.3484692 


3.7797631 


55 


3025 


166375 


7.4161985 


3.8029525 


56 


3136 


175616 


7.4833148 


3.8258624 


57 


3249 


185193 


7.5498344 


3.8485011 


58 


3364 


195112 


7.6157731 


3.8708766 


59 


3481 


205379 


7.6811457 


3.8929965 


60 


3600 


216000 


7.7459667 


3.9148676 


61 


3721 


226981 


7.8102497 


3.9304972 


62 


3844 


238328 


7.8740079 


3.9578915 


63 


3969 


250047 


7.9372539 


3.9790571 


64 


4096 


262144 


8.0000000 


4.0000000 


65 


4225 


274625 


8.0622577 


4.0207256 


66 


4356 


287496 


8.1240384 


4.0412401 


67 


4489 


300763 


8.1853528 


4.0615480 


68 


4624 


314432 


8.2462113 


4.0816551 


69 


4761 


328509 


8.3066239 


4.1015661 


70 


4900 


343000 


8.3666003 


4.1212853 


71 


5041 


357911 


8.4261498 


4.1408178 


72 


5184 


373248 


8.4852814 


4.1601676 


73 


5329 


389017 


8.5440037 


4.1793390 


74 


5476 


405224 


8.6023253 


4.1983364 


75 


5625 


421875 


8.6602540 


4.2171633 


76 


5776 


438976 


8.7177979 


4.2358236 


77 


5929 


456533 


8.7749644 


4.2543210 


78 


6084 


474552 


8.8317609 


4.2726586 


79 


6241 


493039 


8.8881944 


4.2908404 


80 


6400 


512000 


8.9442719 


4.3088695 


81 


6561 


531441 


9.0000000 


4.3267487 


82 


6724 


551368 


9.0553851 


4.3444815 


83 


6889 


571787 


9.1104336 


4.3620707 


84 


7056 


592704 


9.1651514 


4.3795191 



TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 247 



dumber. 


Square. 


Cube. 


Square Root. 


Cube Root. 


85 


7225 


614125 


9.2195445 


4.3968296 


86 


7396 


636056 


9.2736185 


4.4140049 


87 


7569 


658503 


9.3273791 


4.4310476 


88 


7744 


681472 


9.3808315 


4.4479602 


89 


7921 


704969 


9.4339811 


4.4647451 


90 


8100 


729000 


9.4868330 


4.4814047 


91 


8281 


753571 


9.5393920 


4.4979414 


92 


8464 


778688 


9.5916630 


4.5143574 


93 


8649 


804357 


9.6436508 


4.5306549 


94 


8836 


830584 


9.6953597 


4.5468359 


95 


9025 


857374 


9.7467943 


4.5629026 


96 


9216 


884736 


9.7979590 


4.5788570 


97 


9409 


912673 


9.8488578 


4.5947009 


98 


9604 


941192 


9.8994949 


4.6104363 


99 


9801 


970299 


9.9498744 


4.6260650 


100 


10000 


1000000 


10.0000000 


4.6415888 


101 


10201 


1030301 


10.0498756 


4.6570095 


102 


10404 


1061208 


10.0995049 


4.6723287 


103 


10609 


1092727 


10.1488916 


4.6875482 


104 


10816 


1124864 


10.1980390 


4.7026694 


105 


11025 


1157625 


10.2469508 


4.7176940 


106 


11236 


1191016 


10.2956301 


4.7326235 


107 


11449 


1225043 


10.3440804 


4.7474594 


108 


11664 


1259712 


10.3923048 


4.7622032 


. 109 


11881 


1295029 


10.4403065 


4.7768562 


110 


12100 


1331000 


10.4880885 


4.7914199 


111 


12321 


1367631 


10.5356538 


4.8058995 


112 


12544 


1404928 


10.5830052 


4.8202845 


113 


12769 


1442897 


10.6301458 


4.8345881 


114 


12996 


1481544 


10.6770783 


4.8488076 


115 


13225 


1520875 


10.7238053 


4.8629442 


116 


13456 


1560896 


10.7703296 


4.8769990 


117 


13689 


1601613 


10.8166538 


4.8909732 


118 


13924 


1643032 


10.8627805 


4.9048681 


119 


14161 


1685159 


10.9087121 


4.9186847 


120 


14400 


1728000 


10.9544512 


4.9324242 


121 


14641 


1771561 


11.0000000 


4.9460874 


122 


14884 


1815848 


11.0453610 


4.9596757 


123 


15129 


1860867 


11.0905365 


4.9731898 


124 


15376 


1906624 


11.1355287 


4.9866310 


125 


15625 


1953125 


11.1803399 


5.0000000 


126 


15876 


2000376 


11.2249722 


5.0132979 



248 TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


127 


16129 


2048383 


11.2694277 


5.0265257 


128 


16384 


2097152 


11.3137085 


5.0396842 • 


129 


16641 


2146689 


11.3578167 


5.0527743 


130 


16900 


2197000 


11.4017543 


5.0657970 


131 


17161 


2248091 


11.4455231 


5.0787531 


132 


17424 


2299968 


11.4891253 


5.0916434 


133 


17689 


2352637 


11.5325626 


5.1044687 


134 


17956 


2406104 


11.5758369 


5.1172299 


135 


. 18225 


2460375 


11.6189500 


5.1299278 


136 


18496 


2515457 


11.6619038 


5.1425632 


137 


18769 


2571353 


11.7046999 


5.1551367 


138 


19044 


2628072 


11.7473444 


5.1676493 


139 


19321 


2685619 


11.7898261 


5.1801015 


140 


19600 


2744000 


11.8321596 


5.1924941 


141 


19881 


2803221 


11.8743421 


5.2048279 


142 


20164 


2863288 


11.9163753 


5.2171034 


143 


20449 


2924207 


11.9582607 


5.2293215 


144 


20736 


2985984 


12.0000000 


5.2414828 


145 


21025 


3048625 


12.0415946 


5.2535879 


146 


21316 


3112136 


12.0830460 


5.2656374 


147 


21609 


3176523 


12.1243557 


5.2776321 


148 


21904 


3241792 


12.1655251 


5.2895725 


149 


22201 


3307949 


12.2065556 


5.3014592 


150 


22500 


3375000 


12.2474487 


5.3132928 


151 


22801 


3442951 


12.2882057 


5.3250740. 


152 


23104 


3511008 


12.3288280 


5.3368033 


153 


23409 


3581577 


12.3693169 


5.3484812 


154 


23716 


3652264 


12.4096736 


5.3601084 


155 


24025 


3723875 


12.4498996 


5.3716854 


156 


24336 


3796416 


12.4899960 


5.3832126 


157 


24649 


3869893 


12.5299641 


5.3946907 


158 


24964 


3944312 


125698051 


5.4061202 


159 


25281 


4019679 


12.6095202 


5.4175015 


160 


25600 


4096000 


12.6491106 


5.4288352 


161 


25921 


4173281 


12.6885775 


5.4401218 


162 


26244 


4251528 


12.7279221 


5.4513618 


163 


26569 


4330747 


12.7671453 


5.4625556 


164 


26896 


4410944 


12.8062485 


5.4737037 


165 


27225 


4492125 


12.8452326 


5.4848066 


166 


27556 - 


4574296 


12.8840987 


5.4958647 


167 


27889 


4657463 


12.9228480 


5.5068784 


168 


28224 


4741632 


12.9614814 


5.5178484 



TABLE OF SQUARES, CUBES, SQUARE AND CUBE BOOTS. 249 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


169 


28561 


4826809 


13.0000000 


5.5287748 


170 


28900 


4913000 


13.0384048 


5.5396583 


171 


29241 


5000211 


13.0766968 


5.5504991 


172 


29584 


5088448 


13.1148770 


5.5612978 


173 


29929 


5177717 


13.1529464 


5.5 720546 


174 


30276 


5268024 


13.1909060 


5.5827702 


175 


30625 


5359375 


13.2287566 


5.5934447 


176 


30976 


5451776 


13.2664992 


5.6040787 


177 


31329 


5545233 


13.3041347 


5.6146724 


178 


31684 


5639752 


13.3416641 


5.6252263 


179 


32041 


5735339 


13.3790882 


5.6357408 


180 


32400 


5832000 


13.4164079 


5.6462162 


181 


32761 


5929741 


13.4536240 


5.6566528 


182 


33124 


6028568 


13.4907376 


5.6670511 


183 . 


33489 


6128487 


13.5277493 


5.6774114 


184 


33856 


6229504 


13.5646600 


5.6877340 


185 


34225 


6331625 


13.6014705 


5.6980192 


186 


34596 


6434856 


13.6381817 


5.7082675 


187 


34969 


6539203 


13.6747943 


5.7184791 


188 


35344 


6644672 


13.7113092 


5.7286543 


189 


35721 


6751269 


13.7477271 


5.7387936 


190 


36100 


6859000 


13.7840488 


5.7488971 


191 


36481 


6967871 


13.8202750 


5.7589652 


192 


36864 


7077888 


13.8564065 


5.7689982 


193 


37249 


7189057 


13.8924400 


5.7789966 


194 


37636 


7301384 


13.9283883 


5.7889604 


195 


38025 


7414875 


13.9642400 


5.7988900 


196 


38416 


7529536 


14.0000000 


5.8087857 


197 


38809 


7645373 


14.0356688 


5.8186479 


198 


39204 


7762392 


14.0712473 


5.8284867 


199 


39601 


7880599 


14.1067360 


5.8382725 


200 


40000 


8000000 


14.1421356 


5.8480355 


201 


40401 


8120601 


14.1774469 


5.8577660 


202 


40804 


8242408 


14.2126704 


5-8674673 


203 


41209 


8365427 


14.2478068 


5.8771307 


204 


41616 


8489664 


14.2828569 


5.8867653 


205 


42025 


8615125 


14.3178211 


5.8963685 


206 


42436 


8744816 


14.3527001 


5.9059406 


207 


42849 


8869743 


14.3874946 


5.9154817 


208 


43264 


8998912 


14.4222051 


5.9249921 


209 


43681 


9129329 


14.4568323 


5.9344721 


210 


• 44100 


9261000 


14.4913767 


5.9439220 



250 TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


211 


44521 


9393931 


14.5258390 


5.9533418 


212 


44944 


9528128 


14.5602198 


5.9627320 


213 


45369 


9663597 


14.5945195 


5.9720926 


214 


45796 


9800344 


14.6287388 


5.9814241 


215 


46225 


9938375 


14.6628783 


5.9907264 


216' 


46656 


10077696 


14.6969385 


6.0000000' 


217 


47089 


10218313 


14.7309199 


6.0092450 


218 


47524 


10360232 


14.7648231 


6.0184617 


219 


47961 


10503459 


14.7986486 


6.0276502 


220 


48400 


10648000 


14.8323970 


6.0368107 


221 


48841 


10793861 


14.8660687 


6.0459435 


222 


49284 


10941048 


14.8996644 


6.0550489 


223 


49729 


11089567 


14.9331845 


6.0641270 


224 


50176 


11239424 


14.9666295 


6.0731779 


225 


50625 


11390625 


15.0000000 


6.0822020 


226 


51076 


11543176 


15.0332964 


6.0911994 


227 


51529 


11697083 


15.0665192 


6.1001702 


228 


51984 


11852352 


15.0996689 


6.1091147 


229 


52441 


12008989 


15.1327460 


6.1180332 


230 


52900 


12167000 


15.1657509 


6.1269257 


231 


53361 


12326391 


15.1986842 


6.1357924 


232 


53824 


12487168 


15.2315462 


6.1446337 


233 


54289 


12649337 


15.2643375 


6.1534495 


234 


54756 


12812904 


15.2970585 


6.1622401 


235 


55225 


12977875 


15.3297097 


6.1710058 


236 


55696 


13144256 


15.3622915 


6.1797466 


237 


56169 


13312053 


15.3948043 


6.1884628 


238 


56644 


13481272 


15.4272486 


6.1971544 


239 


57121 


13651919 


15.4596248 


6.2058218 


240 


57600 


13824000 


15.4919334 


6.2144650 


241 


58081 


13997521 


15.5241747 


6.2230843 


242 


58564 


14172488 


15.5563492 


6.2316797 


243 


59049 


14348907 


15.5884573 


6.2402515 


244 


59536 


14526784 


15.6204994 


6.2487998 


245 


60025 


14706125 


15.6524758 


6.2573248 


246 


60516 


14886936 


15.6843871 


6.2658266 


247 


61009 


15069223 


15.7162336 


6.2743054 


248 


61504 


15252992 


15.7480157 


6.2827613 


249 


62001 


15438249 


15.7797338 


6.2911946 


250 


62500 


15625000 


15.8113883 


6.2996053 


251 


63001 


15813251 


15.8429795 


6.3079935 


252 


63504 


16003008 


15.8745079 


6.3163596 



TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 251 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


253 


64009 


16194277 


15.9059737 


6.3247035 


254 


64516 


16387064 


15.9373775 


6.3330256 


255 


65025 


16581375 


15.9687194 


6.3413257 


256 


65536 


16777216 


16.0000000 


6.3496042 


257 


66049 


16974593 


16.0312195 


6.3578611 


258 


66564 


17173512 


16.0623784 


6.3660968 


259 


67081 


17373979 


16.0934769 


6.3743111 


260 


67600 


17576000 


16.1245155 


6.3825043 


261 


68121 


17779581 


16.1554944 


6.3906765 


262 


68644 


17984728 


16.1864141 


6.3988279 


263 


69169 


18191447 


16.2172747 


6.4069585 


264 


69696 


18399744 


16.2480768 


6.4150687 


265 


70225 


18609625 


16.2788206 


6.4231583 


266 


70756 


18821096 


16.3095064 


6.4312276 


267 


71289 


19034163 


16.3401346 


6.4392767 


268 


71824 


19248832 


16.3707055 


6.4473057 


269 


72361 


19465109 


16.4012195 


6.4553148 


270 


72900 


19683000 


16.4316767 


6.4633041 


271 


73441 


19902511 


16.4620776 


6.4712736 


272 


73984 


20123648 


16.4924225 


6.4792236 


273 


74529 


20346417 


16.5227116 


6.4871541 


274 


75076 


20570824 


16.5529454 


6.4950653 


275 


75625 


20796875 


16.5831240 


6.5029572 


276 


76176 


21024576 


16.6*32477 


6.5108300 


277 


76729 


21253933 


16.6433170 


6.5186839 


278 


77284 


21484952 


16.6783320 


6.5265189 


279 


77841 


21717639- 


16.7032931 


6.5343351 


280 


78400 


21952000 


16.7332005 


6.5421326 


281 


78961 


22188041 


16.7630546 


6.5499116 


282 


79524 


22425768 


16.7928556 


6.5576722 


283 


80089 


22665187 


16.8226038 


6.5654144 


284 


80656 


22906304 


16.8522995 


6.5731385 


285 


81225 


23149125 


16.8819430 


6.5808443 


286 


81796 


23393656 


16.9115345 


6.5885323 


287 


82369 


.23639903 


16.9410743 


6.5962023 


288 


82944 


23887872 


16.9705627 


6.6038545 


289 


83521 


24137569 


17.0000000 


6.6114890 


290 


84100 


24389000 


17.0293864 


6.6191060 


291 


84681 


24642171 


17.0587221 


6.6267054 


292 


85264 


24897088 


17.0880075 


6.6342874 


293 


85849 


25153757 


17.1172428 


6.6418522 


294 


86436 


25412184 


17.1464282 


6.6493998 



252 TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


295 


87025 


25672375 


17,1755640 


6.6569302 


296 


87616 


25934336 


17.2046505 


6.6644437 


297 


88209 


26198073 


17.2336879 


6.6719403 


298 


88804 


26463592 


17.2626765 


6.6794200 


299 


89401 


26730899 


17.2916165 


6.6868831 


300 


90000 


27000000 


17.3205081 


6.6943295 


301 


90G01 


27270901 


17.3493516 


6.7017593 


302 


91204 


27543608 


17.3781472 


6.7091729 


303 


91809 


27818127 


17.4068952 


6.7165700 


304 


92416 


28094464 


17.4355958 


6.7239508 


305 


93025 


28372625 


17.4642492 


6.7313155 


306 


93636 


28652616 


17.4928557 


6.7386643 


307 


94249 


28934443 


17.5214155 


6.7459967 


308 


94864 


29218112 


17.5499288 


6.7533134 


309 


95481 


29503629 


17.5783958 


6.7606143 


310 


96100 


29791000 


17.6068169 


6.7678995 


311 


96721 


30080231 


17.6351921 


6.7751690 


312 


97344 


30371328 


17.6635217 


6.7824229 


313 


97969 


30664297 


17.6918060 


6.7896613 


.314 


98596 


30959144 


17.7200451 


6.7968844 


315 


99225 


31255875 


17.7482393 


6.8040921 


316 


99856 


31554496 


17.7763888 


6.8112847 


317 


100489 


31855013 


17.8044938 


6.8184620 


318 


101124 ' 


32157432 


17.8325545 


6.8256242 


319 


101761 


32461759 


17.8605711 


6.8327714 


320 


102400 


32768000 


17.8885438 


6.8399037 


321 


103041 


33076161 


17.9164729 


6.8470213 


322 


103684 


33386248 


17.9443584 


6.8541240 


323 


104329 


33698267 


17.9722008 


6.8612120 


324 


104976 


34012224 


18.0000000 


6.8682855 


325 


105625 


34328125 


18.0277564 


6.8753433 


326 


106276 


34645976 


18.0554701 


6.8823888 


327 


106929 


34965783 


18.0831413 


6.8894188 


328 


107584 


35287552 


18.1107703 


6.8964345 


329 


108241 


35611289 


18.1383571 


6.9034359 


330 


108900 


35937000 


18.1659021 


6.9104232 


331 


109561 


36264691 


18.1934054 


6.9173964 


332 


110224 


36594368 


18.2208672 


6. 9243556 


333 


110889 


36926037 


18.2482876 


6.9313088 


334 


111556 


,37259704 


18.2756669 


6.9382321 


335 


112225 


37595375 


18.3030052 


6.9451496 


336 


112896 


37933056 


18.3303028 


6.9520533 



TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 253 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


337 


113569 


38272753 


18.3575598 


6.9589434 


338 


114244 


38614472 


18.3847763 


6.9658198 


339 


114921 


38958219 


18.4119526 


6.9726826 


340 


115600 


39304000 


18.4390889 


6.9795321 


341 


116281 


39651821 


18.4661853 


6.9863681 


342 


116964 


40001688 


18.4932420 


6.9931906 


343 


117649 


40353607 


18.5202592 


7.0000000 


344 


118336 


40707584 


18.5472370 


7.0067962 


345 


119025 


41063625 


18.5741756 


7.0135791 


346 


119716 


41421736 


18.6010752 


7.0203490 


347 


120409 


41781923 


18.6279360 


7.0271058 


348 


121104 


42144192 


18.6547581 


7.0338497 


349 


121801 


42508549 


18.6815417 


7.0405860 


350 


122500 


42875000 


18.7082869 


7.0472987 


351 


123201 


43243551 


18.7349940 


7.0540041 


352 


123904 


43614208 


18.7616630 


7.0606967 


353 


124609 


43986977 


18.7882942 


7.0673767 


354 


125316 


44361864 


18.8148877 


7.0740440 


355 


126025 


44738875 


18.8414437 


7.0806988 


. 356 


126736 


45118016 


18.8679623 


7.0873411 


357 


127449 


45499293 


' 18.8944436 


7.0939709 


358 


128164 


45882712 


18.9208879 


7.1005885 


359 


128881 


46268279 


18.9472953 


7.1071937 


360 


129600 


46656000 


18.9736660 


7.1137866 


361 


130321 


47045881 


19.0000000 


7.1203674 


362 


131044 


47437928 


19.0262976 


7.1269360 


363 


131769 


47832147 


19.0525589 


7.1334925 


364 


132496 


48228544 


19.0787840 


7.1400370 


. 365 


133225 


48627125 


19.1049732 


7.1465695 


366 


133956 


49027896 


19.1311265 


7.1530901 


367 


134689 


49430863 


19.1572441 


7.1595988 


368 


135424 


49836032 


19.1833261 


7.1660957 


369 


136161 


50243409 


19.2093727 


7.1725809 


370 


136900 


50653000 


19.2353841 


7.1790544 


371 


137641 


51064811 


19.2613603 


7.1855162 


372 


138384 


51478848 


19.2873015 


7.1919663 


373 


139129 


51895117 


19.3132079 


7.1984050 


374 


139876 


52313624 


19.3390796 


7.2048322 


375 


140625 


52734375 


19.3649167 


7.2112478 


376 


141376 


53157376 


19.3907194 


7.2176522 


377 


142129 


53582633 


19.4164878 


7.2240450 


378 


142884 


54010152 


19.4422221 


7.2304268 



22 



254 TABLE OP SQUARES, CUBES, SQUARE AND CUBE ROOTS. 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


379 


143641 


54439939 


19.4679223 


7.2367972 


380 


144400 


54872000 


19.4935887 


7.2431565 


381 


145161 


55306341 


19.5192213 


7.2495045 


382 


145924 


55742968 


19.5448203 


7.2558415 


383 


146689 


56181887 


19.5703858 


7.2621675 


384 


147456 


56623104 


19.5959179 


7.2684824 


385 


148225 


57066625 


19.6214169 


7.2747864 


386 


148996 


57512456 


19.6468827 


7.2810794 


387 


149769 


57960603 


19.6723156 


7.2873617 


388 


150544 


58411072 


19.6977156 


7.2936330 


389 


151321 


58863869 


19.7230829 


7.2998936 


390 


152100 


59319000 


19.7484177 


7.3061436 


391 


152881 


59776471 


19.7737199 


7.3123828 


392 


153664 


60236288 


19.7989899 


7.3186114 


393 


154449 


60698457 


19.8242276 


7.3248295 


394 


155236 


61162984 


19.8494332 


7.3310369 


395 


156025 


61629875 


19.8746069 


7.3372339 


396 


156816 


62099136 


19.8997487 


7.3434205 


397 


157609 


62570773 


19.9248588 


7.3495966 


398 


158404 


63044792 


19.9499373 


7.3557624 


399 


159201 


63521199 


19.9749844 


7.3619178 


400 


160000 


64000000 


20.0000000 


7.3680630 


401 


160801 


64481201 


20.0249844 


7.3741979 


402 


161604 


64964808 


20.0499377 


7.3803227 


403 


162409 


65450827 


20.0748599 


7.3864373 


404 


163216 


65939264 


20.0997512 


7.3925418 


405 


164025 


66430125 


20.1246118 


; 7.3986363 


406 


164836 


66923416 


20.1494417 


7.4047206 


407 


165649 


67419143 


20.1742410 


7.4107950 


408 


166464 


67917312 


20.1990099 


7.4168595 


409 


167281 


68417929 


20.2237484 


7.4229142 


410 


168100 


68921000 


20.2484567 


7.4289589 


411 


168921 


69426531 


20.2731349 


7.4349938 


412 


169744 


69934528 


20.2977831 


7.4410189 


413 


170569 


70444997 


20.3224014 


7.4470343 


414 


171396 


70957944 


20.3469899 


7.4530399 


415 


172225 


71473375 


20.3715488 


7.4590359 


416 


173056 


71991296 


20.3960781 


7.4650223 


417 


173889 


72511713 


20.4205779 


7.4709991 


418 


174724 


73034632 


20.4450483 


7.4769664 


419 


175561 


73560059 


20.4694895 


7.4829242 


420 


176400 


74088000 


20.4939015 


7.4888724 



TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 255 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


421 


177241 


74618461 


20.5182845 


7.4948113 


422 


178084 


75151448 


20.5426386 


7.5007406 


423 


178929 


75686967 


20.5669638 


7.5066607 


424 


179776 


76225024 


20.5912603 


7.5125715 


425 


180625 


76765625 


20.6155281 


7.5184730 


426 


181476 


77308776 


20.6397674 


.7.5243652 


427 


182329 


77854483 


20.6639783 


7.5302482 


428 


183184 


78402752 


20.6881609 


7.5361221 


429 


184041 


78953589 


20.7123152 


7.5419867 


430 


184900 


79507000 


20.7364414 


7.5478423 


431 


185761 


80062991 


20.7605395 


7.5536888 


432 


186624 


80621568 


20.7846097 


7.5595263 


433 


187489 


81182737 


20.8086520 


7.5653548 


434 


188356 


81746504 


20.8326667 


7.5711743 


435 


189225 


82312875 


20.8566536 


7.5769849 


436 


190096 


82881856 


20.8806130 


7.5827865 


437 


190969 


83453453 


20.9045450 


7.5885793 


438 


191844 


84027672 


20.9284495 


7.5943633 


439 


192721 


84604519 


20.9523268 


7.6001385 


440 


193600 


85184000 


20.9761770 


7.6059049 


441 


194481 


85766121 


21.0000000 


7.6116626 


442 


195364 


86350888 


21.0237960 


7.6174116 


443 


196249 


86938307 


21.0475652 


7.6231519 


444 


197136 


87528384 


21.0713075 


7.6288837 


445 


198025 


88121125 


21.0950231 


7.6346067 


446 


198916 


88716536 


21.1187121 


7.6403213 


447 


19.9809 


89314623 


21.1423745 


7.6460272 


448 


200704 


89915392 


21.1660105 


7.6517247 


449 


201601 


90518849 


21.1896201 


7.6574138 


450 


202500 


91125000 


21.2132034 


7.6630943 


451 


203401 


91733851 


21.2367606 


7.6687665 


452 


204304 


92345408 


21.2602916 


7.6 744303 


453 


205209 


92959677 


21.2837967 


7.6800857 


454 


206116 


93576664 


21.3072758 


7.6857328 


455 


207025 


94196375 


21.3307290 


7.6913717 


456 


207936 


94818816 


21.3541565 


7.6970023 


457 


208849 


95443993 


21.3775583 


7.7026246 


458 


209764 


96071912 


21.4009346 


7.7082388 


459 


210681 


96702579 


21.4242853 


7.7138448 


460 


211600 


97336000 


21.4476106 


7.7194426 


461 


212521 


97972181 


21.4709106 


7.7250325 


462 


213444 


98611128 


21.4941853 


7.7306141 



256 TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 



Number. 


Square. 


Cube. 


„ Square Root. 


Cube Root. 


463 


214369 


99252847 


21.5174348 


7.7361877 


464 


215296 


99897344 


21.5406592 


7.7417532 


465 


216225 


100544625 


21.5638587 


7.7473109 


466 


217156 


101194696 


21.5870331 


7.7528606 


467 


218089 


101847563 


21.6101828 


7.7584023 


468 


219024 


102503232 


21.6333077 


7.7639361 ' 


469 


219961 


103161709 


21.6564078 


7.7694620 


470 


220900 


103823000 


21.6794834 


7.7749801 


471 


221841 


104487111 


21.7025344 


7.7804904 


472 


222784 


105154048 


21.7255610 


7.7859928 


473 


223729 


105823817 


21.7485632 


7.7914875 


474 


2246 76 


106496424 


21.7715411 


7.7969745 


475 


225625 


107171875 


21.7944947 


7.8024538 


476 


226576 


107850176 


21.8174242 


7.8079254 


477 


227529 


108531333 


21.8403297 


7.8133892 


" 478 


228484 


109215352 


21.8632111 


7.8188456 


479 


229441 


109902239 


21.8860686 


7.8242942 


480 


230400 


110592000 


21.9089023 


7.8297353 


481 


231361 


111284641 


21.9317122 


7.8351688 


482 


232324 


111980168 


21.9544984 


7.8405949 


483 


233289 


112678587 


21.9772610 


7.8460134 


484 


234256 


113379904 


22.0000000 


7.8514244 


485 


235225 


114084125 


22.0227155 


7.8568281 


486 


236196 


114791256 


22.0454077 


7.8622242 


487 


237169 


115501303 


22.0680765 


7.8676130 


488 


238144 


116214272 


22.0907220 


7.8729944 


489 


239121 


116930169 


22.1133444 , 


7.8783684 


490 


240100 


117649000 


22.1359436 


7.8837352 


491 


241081 


118370771 


22.1585198 


7.8890946 


492 


242064 


119095488 


22.1810730 


7.8944468 


493 


243049 


119823157 


22.2036033 


7.8997917 


494 


244036 


120553784 


22.2261108 


7.9051294 


495 


245025 


121287375 


22.2485955 


7.9104599 


496 


246016 


122023936 


22.2710575 


7.9157832 


497 


247009 


122763473 


22.2934968 


7.9210994 


498 


248004 


123505992 


22.3159136 


7.9264085 


499 


249001 


124251499 


22.3383079 


7.9317104 


500 


250000 


125000000 


22.3606798 


7.9370053 


501 


251001 


125751501 


22.3830293 


7.9422931 


502 


252004 


126506008 


22.4053565 


7.9475739 


503 


253009 


127263527 


22.4276615 


7.9528477 


504 


254016 


128024064 


22.4499443 


7.9581144 



TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 257 



Number. 


Square. 


Cube. 


Square Eoot. 


Cube Eoot. 


505 • 


255025 


128787625 


22,4722051 


7.9633743 


506 


25G036 


129554216 


22.4944438 


7.9686271 


507 


257049 


130323843 


22.5166605 


7.9738731 


508 


258064 


131096512 


22.5388553 


7.9791122 


509 


259081 


131872229 


22.5610283 


7.9843444 


510 


260100 


132651000 


22.5831796 


7.9895697 


511 


261121 


133432831 


22.6053091 


7.9947883 


512 


262144 


134217728 


22.6274170 


8.0000000 


513 


263169 


135005697 


22.6495033 


8.0052049 


514 


264196 


135796744 


22.6715681 


8.0104032 


515 


265225 


136590875 


22.6936114 


8.0155946 


516 


266256 


137388096 


22.7156334 


8.0207794 


517 


267289 


138188413 


22.7376341 


8:0259574 


518 


268324 


138991832 


22.7596134 


8.0311287 


519 


269361 


139798359 


22.7815715 


8.0362935 


520 


270400 


140608000 


22.8035085 


8.0414515 


521 


271441 


141420761 


22.8254244 


8.0466030 


522 


272484 


142236648 


22.8473193 


8.0517479 


523 


273529 


143055667 


22.8691933 


8.0568862 


524 


274576 


143877824 


22.8910463 


8.0620180 


525 


275625 


144703125 


22.9128785 


8.0671432 


526 


276676 


145531576 


22.9346899 


8.0722620 


527 


277729 


146363183 


22.9564806 


8.0773743 


528 


278784 


147197952 


22.9782506 


8.0824800 


529 


279841 


148035889 


-23.0000000 


8.0875794 


530 


280900 


148877000 


23.0217289 


8.0926723 


531 


281961 


149721291 


23.0434372 


8.0977589 


532 


283024 


150568768 


23.0651252 


8.1028390 


533 


284089 


151419437 


23.0867928 


8.1079128 


534 


285156 


152273304 


23.1084400 


8.1129803 


535 


286225 


153130375 


23.1300670 


8.1180414 


536 


287296 


153990656 


23.1516738 


8.1230962 


537 


288369 


154854153 


23.1732605 


8.1281447 


538 


289444 


155720872 


23.1948270 


8.1331870 


539 


290521 


156590819 


23.2163735 


8.1382230 


540 


291600 


157464000 


23.2379001 


8.1432529 


541 


292681 


158340421 


23.2594067 


8.1482765 


542 


293764 


159220088 


23.2808935 


8.1532939 


543 


294849 


160103007 


23.3023604 


8.1583051 


544 


295936 


160989184 


23.3238016 


8.1633102 


545 


297025 


161878625 


23.3452351 


8.1683092 


546 


298116" 


162771336 


23.3666429 


8.1733020 



22* 



258 TABLE OE SQUARES, CUBES, SQUARE AND CUBE ROOTS. 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


547 


299209 


163667323 


23.3880311 


8.1782888 


548 


300304 


164566592 


23.4093998 


8.1832695 


549 


301401 


165469149 


23.4307490 


8.1882441 


550 


302500 


166375000 


23.4520788 


8.1932127 


551 


303601 


167284151 


23.4733892 


8.1981753 


552 


304704 


168196608 


23.4946802 


8.2031319 


553 


305809 


169112377 


23.5159520 


8.2080825 


554 


306916 


170031464 


23.5372046 


8.2130271 


555 


308025 


170953875 


23.5584380 


8.2179657 


556 


309136 


171879616 


23.5796522 


8.2228985 


557 


310249 


172808693 


23.6008474 


8.2278254 


558 


311364 


173741112 


23.6220236 


8.2327463 


559 


312481 


174676879 


23.6431808 


8.2376614 


560 


313600 


175616000 


23.6643191 


8.2425706 


561 


314721 


176558481 


23.6854386 


8.2474740 


562 


315844 


177504328 


23.7065392 


8.2523715 


563 


316969 


178453547 


23.7276210 


8.2572635 


564 


318096 


179406144 


23.7486842 


8.2621492 


565 


319225 


180362125 


23.7697286 


8.2670294 


566 


320356 


181321496 


23.7907545 


8.2719039 


567 


321489 


182284263 


23.8117618 


8.2767726 


568 


322624 


183250432 


23.8327506 


8.2816255 


569 


323761 


184220009 


23.8537209 


8.2864928 


570 


324900 


185193000 


23.8746728 


8.2913444 


571 


326041 


186169411 


23.8956063 


8.2961903 


572 


327184 


187149248 


23.9165215 


8.3010304 


573 


328329 


188132517 


23.9374184 


8.3058651 


574 


329476 


189119224 


23.9582971 


8.3106941 


575 


330625 


190109375 


23.9791576 


8.3155175 


576 


331776 


191102976 


24.0000000 


8.3203353 


577 


332929 


192100033 


24.0208243 


8.3251475 


578 


334084 


193100552 


24.0416306 


8.3299542 


579 


335241 


194104539 


24.0624188 


8.3347553 


580 


336400 


195112000 


24.0831891 


8.3395509 


581 


337561 


196122941 


24.1039416 


8.3443410 


582 


338724 


197137368 


22.1246762 


8.3491256 


583 


339889 


198155287 


24.1453929 


8.3539047 


584 


341056 


199176704 


24.1660919 


8.3586784 


585 


342225 


200201625 


24.1867732 


8.3634466 


586 


343396 


201230056 


24.2074369 


8.3682095 


587 


344569 


202262003 


24.2280829 


8.3729668 


588 


345744 


203297472 


24.2487113 


8.3777188 



TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 259 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


589 


346921 


204336469 


24.2693222 


8.3824653 


590 


348100 


205379000 


24.2899156 


8.3872065 


591 


349281 


206425071 


24.3104996 


8.3919428 


592 


350464 


207474688 


24.3310501 


8.3966729 


593 


351649 


208527857 


24.3515913 


8.4013981 


594 


352836 


209584584 


24.3721152 


8.4061180 


595 


354025 


210644875 


24.3926218 


8.4108326 


596 


355216 


211708736 


24.4131112 


8.4155419 


597 


356409 


212776173 


24.4335834 


8.4202460 


598 


357604 


213847192 


24.4540385 


8.4249448 


599 


358801 


214921799 


24.4744765 


8.4296383 


600 


360000 


216000000 


24.49489 74 


8.4343267 


601 


361201 


217081801 


24.5153013 


8.4390098 


602 


362404 


218167208 


24.5356883 


8.4436877 


603 


363609 


219256227 


24.5560583 


8.4483605 


604 


364816 


220348864 


24.5764115 


8.4530280 


605 


366025 


221445125 


24.5967478 


8.4576906 


606 


367236 


222545016 


24.6170673 


8.4623479 


607 


368449 


223648543 


24.6373700 


8.4670001 


608 


369664 


224755712 


24.6576560 


8.4716471 


609 


370881 


225866529 


24.6779254 


8.4762892 


610 


372100 


226981000 


24.6981781 


8.4809261 


611 


373321 


228099131 


24.7184142 


8.4855579 


612 


374544 


229220928 


24.7386338 


8.4901848 


613 


375769 


230346397 


24.7588368 


8.4948065 


614 


376996 


231475544 


24.7790234 


8.4994233 


615 


378225 


232608375 


24.7991935 


8.5040350 


616 


379456 


233744896 


24.8193473 


8.5086417 


617 


380689 


234885113 


24.8394847 


8.5132435 


618 


381924 


236029032 


24.8596058 


8.5178403 


619 


383161 


237176659 


24.8797106 


8.5224331 


620 


384400 


238328000 


24.8997992 


8.5270189 


621 


385641 


239483061 


24.9198716 


8.5316009 


622 


386884 


240641848 


24.9399278 


8.5361780 


623 


388129 


241804367 


24.9599679 


8.5407501 


624 


389376 


242970624 


24.9799920 


8.5453173 


625 


390625 


244140625 


25.0000000 


8.5498797 


626 


391876 


245314376 


25.0199920 


8.5544372 


627 


393129 


246491883 


25.0399681 


8.5589899 


628 


394384 


247673152 


25.0599282 


8.5635377 


629 


395641 


248858189 


25.0798724 


8.5680807 


630 


396900 


250047000 


25.0998008 


8.5726189 



260 TABLE OE SQUARES, CUBES, SQUARE AND CUBE ROOTS. 



Kurnber. 


Square. 


Cube. 


Square Root. 


Cube Root. 


631 


398161 


251239591 


25.1197134 


8.5771523 


632 


399424 


252435968 


25.1396102 


8.5816809 


633 


400689 


253636137 


25.1594913 


8.5862047 


634 


401956 


254840104 


25.1793566 


8.5907238 


• . 635 


403225 


256047875 


25.1992063 


8.5952380 


636 


404496 


257259456 


25.2190404 


8.5997476 


637 


405769 


258474853 


25.2388589 


8.6042525 


638 


407044 


259694072 


25.2586619 


8.6087526 


639 


408321 


260917119 


25.2784493 


8.6132480 


640 


409600 


262144000 


25.2982213 


8.6177388 


641 


410881 


263374721 


25.3179778 


8.6222248 


642 


412164 


264609288 


25.3377189 


8.6267063 


643 


413449 


265847707 


25.3574447 


8.6311830 


644 


414736 


267089984 


25.3771551 


8.6356551 


645 


416025 


268336125 


25.3968502 


8.6401226 


646 


417316 


269586136 


25.4165302 


8.6445855 


647 


418609 


270840023 


25.4361947 


8.6490437 


648 


419904 


272097792 


25.4558441 


8.6534974 


649 


421201 


273359449 


25.4754784 


8.6579465 


650 


422500 


274625000 


25.4950976 


8.6623911 


651 


423801 


275894451 


25.5147013 


8.6668310 


652 


425104 


277167808 


25.5342907 


8.6712665 


653 


426409 


278445077 


25.5538647 


8.6756974 


654 


427716 


279726264 


25.5734237 


8.6801237 


655 


429025 


281011375 


25.5929678 


8.6845456 


656 


430336 


282300416 


25.6124969 


8.6889630 


657 


431649 


283593393 


25.6320112 


8.6933759 


658 


432964 


284890312 


25.6515107 


8.6977843 


659 


434281 


286191179 


25.6709953 


8.7021882 


660 


435600 


287496000 


25.6904652 


8.7065877 


661 


436921 


288804781 


25.7099203 


8.7109827 


6G2 


438244 


290117528 


25.7293607 


8.7153734 


663 


439569 


291434247 


25.7487864 


8.7197596 


664 


440896 


292754944 


25.7681975 


8.7241414 


665 


442225 


294079625 


25.7875939 


8.7285187 


666 


443556 


295408296 


25.8069758 


8.7328918 


667 


444889 


296740963 


25.8263431 


8.7372604 


668 


446224 


298077632 


25.8456960 


8.7416246 


669 


447561 


299418309 


25.8650343 


8.7459846 . 


670 


448900 


300763000 


25.8843582 


8.7503401 


6 71 


450241 


302111711 


25.9036677 


8.7546913 


672 


451584 


303464448 


25.9229628 


8.7590383 



TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 261 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


673 


452929 


304821217 


25.9422435 


8.7633809 


674 


454276 


306182024 


25.9615100 


8.7677192 


675 


455625 


307546875 


25.9807621 


8.7720532 


676 


456976 


308915776 


26.0000000 


8.7763830 


677 


458329 


310288733 


26.0192237 


8.7807084 


678 


459684 


311665752 


26.0384331 


8.7850296 


679 


461041 


313046839 


26.0576284 


8.7893466 


680 


462400 


314432000 


26.0768096 


8.7936593 


681 


463761 


315821241 


26.0959767 


8.7979679 


682 


465124 


317214568 


26.1151297 


8.8022721 


683 


466489 


318611987 


26.1342687 


8.8065722 


684 


467856 


320013504 


26.1533937 


8.8108681 


685 


469225 


321419-125 


26.1725047 


8.8151598 


686 


470596 


322828856 


26.1916017 


8.8194474 


687 


471969 


324242703 


26.2106848 


8.8237307 


688 


473344 


325660672 


26.2297541 


8.8280099 


689 


474721 


327082769 


26.2488095 


8.8322850 


690 


476100 


328509000 


26.2678511 


8.8365559 


691 


477481 


329939371 


26.2868789 


8.8408227 


692 


478864 


331373888 


26.3058929 


8.8450854 


693 


480249 


332812557 


26.3248932 


8.8493440 


694 


481636 


334255384 


26.3438797 


8.8535985 


695 


483025 


335702375 


26.3628527 


8.8578489 


696 


484416 


337153536 


26.3818119 


8.8620952 


697 


485809 


338608873 


26.4007576 


8.8663375 


698 


487204 


340068392 


26.4196896 


8.8705757 


699 


488601 


341532099 


26.4386081 


8.8748099 


700 


490000 


343000000 


26.4575131 


8.8790400 


701 


491401 


344472101 


26.4764046 


8.8832661 ' 


702 


492804 


345948408 


26.4952826 


8.8874882 


703 


494209 


347428927 


26.5141472 


8.8917063 


704 


495616 


348913664 


26.5329983 


8.8959204 


705 


497025 


350402625 


26.5518361 


8.9001304 


706 


498436 


351895816 


26.5706605 


8.9043366 


707 


499849 


• 353393243 


26.5894716 


8.9085387 


708 


501264 


354894912 


26.6082694 


8.9127369 


709 


502681 


356400829 


26.6270539 


8.9169311 


710 


504100 


357911000 


26.6458252 


8.9211214 


711 


505521 


359425431 


26.6645833 


8.9253078 


712 


506944 


360944128 


26.6833281 


8.9294902 


713 


508369 


362467097 


26.7020598 


8.9336687 


714 


509796 


363994344 


26.7207784 


8.9378433 



262 TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


715 


511225 


365525875 


26.7394839 


8.9420140 


716 


512656 


367061696 


26.7581763 


8.9461809 


717 


514089 


368601813 


26.7768557 


8.9503438 


718 


515524 


370146232 


26.7955220 


8.9545029 


719 


516961 


371694959 


26.8141754 


8.9586581 


720 


518400 


373248000 


26.8328157 


8.9628095 


721 


519841 


374805361 


26.8514432 


8.9669570 


722 


521284 


376367048 


26.8700577 


8.9711007 


723 


522729 


377933067 


26.8886593 


8.9752406 


724 


524176 


379503424 


26.9072481 


8.9793766 


725 


525625 


381078125 


26.9258240 


8.9835089 


726 


527076 


382657176 


26.9443872 


8.9876373 


727 


528529 


384240583 


26.9629375 


8.9917620 


728 


529984 


385828352 


26.9814751 


8.9958899 


729 


531441 


387420489 


27.0000000 


9.0000000 


730 


532900 


389017000 


27.0185122 


9.0041134 


731 


534361 


390617891 


27.0370117 


9.0082229 


732 


535824 


392223168 


27.0554985 


9.0123288 


733 


537289 


393832837 


27.0739727 


9.0164309 


734 


538756 


395446904 


27.0924344 


9.0205293 


735 


540225 


397065375 


27.1108834 


9.0246239 


736 


541696 


398688256 


27.1293199 


9.0287149 


737 


543169 


400315553 


27.1477149 


9.0328021 


738 


544644 


401947272. 


27.1661554 


9.0368857 


739 


546121 


403583419 


27.1845544 


9.0409655 


740 


547600 


405224000 


27.2029140 


9.0450419 


741 


549081 


406869021 


27.2213152 


9.0491142 


742 


550564 


408518488 


27.2396769 


9.0531831 


743 


552049 


410172407 


27.2580263 


9.0572482 


744 


553536 


411830784 


27.2763634 


9.0613098 


745 


555025 


413493625 


27.2946881 


9.0653677 


746 


556516 


415160936 


27.3130006 


9.0694220 


747 


558009 


416832723 


27.3313007 


9.0734726 


748 


559504 


418508992 


27.3495887 


9.0775197 


749 


561001 


420189749 


27.3678644 


9.0815631 


750 


562500 


421875000 


27.3861279 


9.0856030 


751 


564001 


423564751 


27.4043792 


9.0896392 


752 


565504 


425259008 


27.4226184 


9.0936719 


753 


567009 


426957777 


27.4408455 


9.0977010 


754 


568516 


428661064 


27.4590604 


9.1017265 


755 


570025 


430368875 


27.4772633 


9.1057485 


756 


571536 


432081216 


27.4954542 


9.1097669 



TABLE OP SQUARES, CUBES, SQUARE AND CUBE ROOTS. 263 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


757 


573049 


433798093 


27.5136330 


9.1137818 


758 


574564 


435519512 


27.5317998 


9.1177931 


759 


576081 


437245479 


27.5499546 


9.1218010 


760 


577600 


438976000 


27.5680975 


9.1258053 


761 


579121 


440711081 


27.5862284 


9.1298061 


762 


580644 


442450728 


27.6043475 


9.1338034 


763 


582169 


444194947 


27.6224546 


9.1377971 


764 


583696 


445943744 


27.6405499 


9.1417875 


765 


585225 


447697125 


27.6586334 


9.1457742 


766 


586756 


449455096 


27.6767050 


9.1497576 


767 


588289 


451217663 


27.6947648 


9.1537375 


768 


589824 


452984832 


27.7128129 


9.1577139 


769 


591361 


454756609 


27.7308492 


9.1616869 


770 


592900 


456533000 


27.7488739 


9.1656565 


771 


594441 


458314011 


27.7668868 


9.1696225 


772 


595984 


460099648 


27.7848880 


9.1735852 


773 


597529 


461889917 


27.8028775 


9.1775445 


774 


599076 


463684824 


27.8208555 


9.1815003 


775 


600625 


465484375 


27.8388218 


9.1854527 


776 


602176 


467288576 


27.8567766 


9.1894018 


777 


603729 


469097433 


27.8747197 


9.1933474 


778 


605284 


470910952 


27.8926514 


9.1972897 


779 


606841 


472729139 


27.9105715 


9,2012286 


780 


608400 


474552000 


27.9284801 


9.2051641 


781 


609961 


476379541 


27.9463772 


9.2090962 


782 


611524 


478211768 


27.9642629 


9.2130250 


783 


613089 


480048687 


27.9821372 


9.2169505 


784 


614656 


481890304 


28.0000000 


9.2208726 


785 


616225 


483736625 


28.0178515 


9.2247914 


786 


617796 


485587656 


28.0356915 


9.2287068 


787 


619369 


487443403 


28.0535203 


9.2326189 


788 


620944 


489303872 


28.0713377 


9.236527 7 


789 


622521 


491169069 


28.0891438 


9.2404333 


790 


624100 


493039000 


28.1069386 


9.2443355 


791 


625681 


494913671 


28.1247222 


9.2482344 


792 


627264 


496793088 


28.1424946 


9.2521300 


793 


628849 


498677257 


28.1602557 


9.2560224 


794 


630436 


500566184 


28.1780056 


9.2599114 


795 


632025 


502459875 


28.1957444 


9.2637973 


796 


633616 


504358336 


28.2134720 


9.2676798 


797 


635209 


506261573 


28.2311884 


9.2715592 


798 


636804 


508169592 


28.2488938 


9.2754352 



264 TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


799 


638401 


510082399 


28.2665881* 


9.2793081 


800 


640000 


512000000 


28.2842712 


9.2831777 


801 


641601 


513922401 


28.3019434 


9.2870444 


802 


643204 


515849608 


28.3196045 


9.2909072 


803 


644809 


517781627 


28.3372546 


9.2947671 


804 


646416 


519718464 


28.3548938 


9.2986239 


805 


648025 


521660125 


28.3725219 


9 3024775 


806 


649636 


523606616 


28.3901391 


9.3063278 


807 


651249 


525557943 


28.4077454 


9.3101750 


808 


652864 


527514112 


28.4253408 


9.3140190 


809 


654481 


529475129 


28.4429253 


9.3178599 


810 


656100 


531441000 „ 


28.4604989 


9.3216975 


811 


657721 


533411731 


28.4780617 


9.3255320 


812 


659344 


535387328 


28.4956137 


9.3293634 


813 


660969 


537367797 


28.5131549 


9.3331916 


814 


662596 


539353144 


28.5306852 


9.3370167 


815 


664225 


541343375 


28.5482048 


9.3408386 


816 


665856 


543338496 


28.5657137 


9.3446575 


817 


66 7489 


545338513 


28.5832119 


9.3484731 


818 


669124 


547343432 


28.6006993 


9.3522857 


819 


670761 


549353259 


28.6181760 


9.3560952 


820 


672400 


551368000 


28.6356421 


9.3599016 


821 


674041 


553387661 


28.6530976 


9.3637049 


822 


675684 


555412248 


28.6705424 


9.3675051 


823 


677329 


557441767 


28.6879716 


9.3713022 


824 


678976 


559476224 


28.7054002 


9.3750963 


825 


680625 


561515625 


28.7228132 


9.3788873 


826 


682276 


563559976 


28.7402157 


9.3826752 


827 


683929 


565609283 


28.7576077 


9.3864600 


828 


685584 


567663552 


28.7749891 


9.3902419 


829 


687241 


569722789 


28.7923601 


9.3940206 


830 


688900 


571787000 


28.8097206 


9.3977964 


831 


690561 


573856191 


28.8270706 


9.4015691 


832 


692224 


575930368 


28.8444102 


9.4053387 


833 


693889 


578009537 


28.8617394 


9.4091054 


834 


695556 


580093704 


28.8790582 


9.4128690 


835 


697225 


582182875 


28.8963666 


9.4166297 


836 


698896 


584277056 


28,9136646 


9.4203873 


837 


700569 


586376253 


28.9309523 


9.4241420 


838 


702244 


588480472 


28.9482297 


9.4278936 


839 


703921 


590589719 


28.9654967 


9.4316423 


840 


705600 


592704000 


28.9827535 


9.4353880 



TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 265 



Kumber. 


Square. 


Cube. 


Square Root, 


Cube Root. 


841 


707281 


594823321 


29.0000000 


9.4391307 


842 


708964 


596947688 


29.0172363 


9.4428704 


843 


710649 


599077107 


29.0344623 


9.4466072 


844 


712336 


601211584 


29.0516781 


9.4503410 


845 


714025 


603351125 


29.0688837 


9.4540719 


846 


715716 


605495736 


29.0860791 


9.4577999 


847 


717409 


607645423 


29.1032644 


9.4615249 


848 


719104 


609800192 


29.1204396 


9.4652470 


849 


720801 


611960049 


29.1376046 


9.4689661 


850 


722500 


614125000 


29.1547595 


9.4726824 


851 


724201 


616295051 


29.1719043 


9.4763957 


852 


725904 


618470208 


29.1890390 


9.4801061 


853 


727609 


620650477 


29.2061637 


9.4838136 


854 


729316 


622835864 


29.2232784 


9.4875182 


855 


731025 


625026375 


29.2403830 


9.4912200 


856 


732736 


627222016 


29.2574777 


9.4949188 


857 


734449 


629422793 


29.2745623 


9.4986147 


858 


736164 


631628712 


29.2916370 


9.5023078 


859 


737881 


633839779 


29.3087018 


9.5059980 


860 


739600 


636056000 


29.3257566 


9.5096854 


861 


741321 


638277381 


29.3428015 


9.5133699 


862 


743044 


640503928 


29.3598365 


9.5170515 


863 


744769 


642735647 


29.3768616 


9.5207303 


864 


746496 


644972544 


29.3938769 


9.5244063 


865 


748225 


647214625 


29.4108823 


9.5280794 


866 


749956 


649461896 


29.4278779 


9.5317497 


867 


751689 


651714363 


29.4448637 


9.5354172 


868 


753424 


653972032 


29.4618397 


9.5390818 


869 


755161 


656234909 


25.4788059 


9.5427437 


870 


756900 


658503000 


29.4957624 


9.5464027 


871 


758641 


660776311 


29.5127091 


9.5500589 


872 


760384 


663054848 


29.5296461 


9.5537123 


873 


762129 


665338617 


29.5465734 


9.5573630 


874 


763876 


667627624 


29.5634910 


9.5610108 


875 


765625 


669921875 


29.5803989 


9.5646559 


876 


767376 


672221376 


29.5972972 


9.5682982 


877 


769129 


674526133 


29.6141858 


9.5719377 


878 


770884 


676836152 


29.6310648 


9.5755745 


879 


772641 


679151439 


29.6479342 


9.5792085 


880 


774400 


681472000 


29.6647939 


9.5828397 


881 


776161 


683797841 


29.6816442 


9.5864682 


882 


777924 


686128968 • 


29.6984848 


. 9.5900937 



23 



266 TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 



Number. 


Square. 


Cube. 


Square Root. 


Cube Root. 


883 


779689 


688465387 


29.7153159 


9.5937169 


884 


781456 


690807104 


29.7321375 


9.5973373 


885 


783225 


693154125 


29.7489496 


9.6009548 


886 


784996 


695506456 


29.7657521 


9.6045696 


887 


786769 


697864103 


29.7825452 


9.6081817 


888 


788544 


700227072 


29.7993289 


9.6117911 


889 


790321 


702595369 


29.8161030 


9.6153977 


890 


7.92100 


704969000 


29.8328678 


9.6190017 


891 


793881 


707347971 


29.8496231 


9.6226030 


892 


795664 


707932288 


29.8663690 


9.6262016 


893 


797449 


712121957 


29.8831056 


9.6297975 


894 


799236 


714516984 


29.8998328 


9.6333907 


895 


801025 


716917375 


29.9165506 


9.6369812 


896 


802816 


719323136 


29.9332591 


9.6405690 


897 


804609 


721734273 


29.9499583 


9.6441542 


898 


806404 


724150792 


29.9666481 


9.6477367 


899 


808201 


726572699 


29.9833287 


9.6513166 


900 


810000 


729000000 


30.0000000 


9.6548938 


901 


811801 


731432701 


30.0166621 


9.6584684 


902 


813604 


733870808 


30.0333148 


9.6620403 


903 


815409 


736314327 


30.0499584 


9.6656096 


904 


817216 


738763264 


30.0665928 


9.6691762 


905 


819025 


741217625 


30.0832179 


9.6727403 


906 


820836 


743677416 


30.0998339 


9.6763017 


907 


822649 


746142643 


30.1164407 


9.6798604 


908 


824464 


748613312 


30.1330383 


9.6834166 


909 


826281 


751089429 


30.1496269 


9.6869701 


910 


828100 


753571000 


30.1662063 


9.6905211 


911 


829921 


756058031 


30.1827765 


9.6940694 


912 


831744 


758550528 


30.1993377 


9.6976151 


913 


833569 


761048497 


30-.2158899 


9.7011583 


914 


835396 


763551944 


30.2324329 


9.7046989 . 


915 


837225 


766060875 


30.2489669 


9.7082369 


916 


839056 


768575296 


30.2654919 


9.7117723 


917 


840889 


771095213 


30.2820079 


9.7153051 


, 918 


842724 


773620632 


30.2985148 


9.7188354 


919 


844561 


776151559 


30.3150128 


9.7223631 


920 


846400 


778688000 


30.3315018 


9.7258883 


921 


848241 


781229961 


30.3479818 


9.7294109 


922 


850084 


783777448 


30.3644529 


9.7329309 


923 


851929 


786330467 


30.3809151 


9.7364484 


924 


$53776 


788889024 


30.3973683 


9.7399634 



TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 267 



Number. 
925 


Square. 


Cube. 


Square Koot. 


Cube Root. 


855625 


791453125 


30.4138127 


9.7434758 


926 


857476 


794022776 


30.4302481 


9.7469857 


927 


859329 


796597983 


30.4466747 


9.7504930 


928 


861184 


799178752 


30.4630924 


9.7539979 


929 


863041 


801765089 


30.4795013 


9.7575002 


930 


864900 


804357000 


30.4959014 


9.7610001 


931 


866761 


806954491 


30.5122926 


9.7644974 


932 


868624 


809557568 


30.5286750 


9.7679922 


933 


870489 


812166237 


30.5450487 


9.7714845 


934 


872356 


814780504 


30.5614136 


9.7749743 


935 


874225 


817400375 


30.5777697 


9.7784616 


936 


876096 


820025856 


30.5941171 


9.7829466 


937 


877969 


822656953 


30.6104557 


9.7854288 


938 


879844 


825293672 


30.6267857 


9.7889087 


939 


881721 


827936019 


30.6431069 


9.7923861 


940 


883600 


830584000 


30.6594194 


9.7958611 


941 


885481 


833237621 


30.6757233 


9.7993336 


942 


887364 


835896888 


30.6920185 


9.8028036 


943 


889249 


838561807 


30.7083051 


9.8062711 


944 


891136 


841232384 


30.7245830 


9.8097362 


945 


893025 


843908625 


30.7408523 


9.8131989 


946 


894916 


846590536 


30.7571130 


9.8166591 


947 


896808 


849278123 


30.7733651 


9.8201169 


948 


898704 


851971392 


30.7896086 


9.8235723 


949 


900601 


854670349 


30.8058436 


9.8270252 


950 


902500 


857375000 


30.8220700 


9.8304757 


951 


9.04401 


860085351 


30.8382879 


9.8339238 


952 


906304 


862801408 


30.8544972 


9.8373695 


953 


908209 


865523177 


30.8706981 


9.8408127 


954 


910116 


868250664 


30.8868904 


9.8442536 


955 


912025 


870983875 


30.9030743 


9.8476920 


956 


913936 


873722816 


30.9192477 


9.8511280 


957 


915849 


876467493 


30.9354166 


9.8545617 


958 


917764 


879217912 


30.9515751 


9.8579929 


959 


919681 


881974079 


30.9677251 


9.8614218 


960 


921600 


884736000 


30.9838668 


9.8648483 


961 


923521 


887503681 


31.0000000 


9.8682724 


962 


925444 


890277128 


31.0161248 


9.8716941 


963 


927369 


893056347 


31.0322413 


9.8751135 


964 


929296 


895841344 


31.0483494 


9.8785305 


965 


931225 


898632125 


31.0644491 


9.8819451 


966 


933156 


901428696 


31.0805405 


9.8853574 



268 TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 



Number. 


Square. 


Cube. 


Square Root. 


- 
Cube Root. 


967 


935089 


904231063 


31.0966236 


9.8887673 


968 


937024 


907039232 


31.1126984 


9.8921749 


969 


938961 


909853209 


31.1287648 


9.8955801 


970 


940900 


912673000 


31.1448230 


9.8989830 


971 


942841 


915498611 


31.1608729 


9.9023835 


972 


944784 


918330048 


31.1769145 


9.9057817 


973 


946729 


921167317 


31.1929479 


9.9091776 


974 


948676 


924010424 


31.2089731 


9.9125712 


975 


950625 


926859375 


31.2249900 


9.9159624 


976 


952576 


929714176 


31.2409987 


9.9193513 


977 


954529 


932574833 


31.2569992 


9.9227379 


978 


556484 


935441352 


31.2729915 


9.9261222 


979 


958441 


938313739 


31.2889757 


9.9295042 


980 


960400 


941192000 


31.3049517 


9.9328839 


981 


962361 


944076141 


31.3209195 


9.9362613 


982 


964324 


946966168 


31.3368792 


9.9396363 


983 


966289 


949862087 


31.3528308 


9.9430092 


• 984 


968256 


952763904 


31.3687743 


9.9463797 


985 


970225 


955671625 


31.3847097 


9.9497479 


986 


972196 


958585256 


31.4006369 


9.9531138 


987 


974169 


961504803 


31.4165561 


9.9564775 


988 


976144 


964430272 


31.4324673 


9.9598389 


989 


978121 


967361669 


31.4483704 


9.9631981 


990 


980100 


970299000 


31.4642654 


9.9665549 


991 


982081 


973242271 


31.4801525 


9.9699095 


992 


984064 


976191488 


31.4960315 


9.9732619 


993 


986049 


979146657 


31.5119025 


9.9766120 


994 


988036 


982107784 


31.5277655 ' 


9.9799599 


995 


990025 


985074875 


31.5436206 


9.9833055 


996 


992016 


988047936 


31.5594677 


9.9866488 


997 


994009 


991026973 


31.5753068 


9.9899900 


998 


996004 


994011992 


31.5911380 


9.9933289 


999 


998001 


997002999 


31.6069613 


9.9966656 


1000 


1000000 


1000000000 


31.6227766 


10.0000000 


1001 


1002001 


1003003001 


31.6385840 


10.0033222 


1002 


1004004 


1006012008 


31.6543836 


10.0066622 


1003 


1006009 


1009027027 


31.6701752 


10.0099899 


1004 


1008016 


1012048064 


31.6859590 


10.0133155 


1005 


1010025 


1015075125 


31.7017349 


10.0166389 


1006 


1012036 


1018108216 


31.7175030 


10.0199601 


1007 


1014049 


1021147343 


31.7332633 


10.0232791 


1008 


1016064 


1024192512 


31.7490157 


10.0265958 
1 



TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 269 



Number. 


Square. 


Cube. 


Square Boot. 


Cube Root. 


1009 


1018081 


1027243729 


31.7647603 


10.0299104 


1010 


1020100 


1030301000 


31.7804972 


10.0332228 


1011 


1022121 


1033364331 


31.7962262 


10.0365330 


1012 


1024144 


1036433728 


31.8119474 


10.0398410 


1013 


1026169 


1039509197 


31.8276609 


10.0431469 


1014 


1028196 


1042590744 


31.8433666 


10.0464506 


1015 


1030225 


1045678375 


31.8590646 


10.0497521 


1016 


1032256 


1048772096 


31.8747549 


10.0530514 


1017 


1034289 


1051871913 


31.8904374 


10.0563485 


1018 


1036324 


1054977832 


31.9061123 


10.0596435 


1019 


1038361 


1058089859 


31.9217794 


10.0629364 


1020 


1040400 


1061208000 


31.9374388 


10.0662271 


1021 


1042441 


1064332261 


31.9530906 


10.0695156 


1022 


1044484 


1067462648 


31.9687347 


10.0728020 


1023 


1046529 


1070599167 


31.9843712 


10.0760863 


1024 


1048576 


1073741824 


32.0000000 


10.0793684 


1025 


1050625 


1076890625 


32.0156212 


10.0826484 


1026 


10526 76 


1080045576 


32.0312348 


10.0859262 


1027 


1054729 


1083206683 


32.0468407 


10.0892019 


1028 


1056784 


1086373952 


32.0624391 


10.0924755 


1029 


1058841 


1089547389 


32.0780298 


10.0957469 


1030 


1060900 


1092727000 


32.0936131 


10.0990163 


1031 


1062961 


1095912791 


32.1091887 


10.1022835 


1032 


1065024 


.1099104768 


32.1247568 


10.1055487 


1033 


1067089 


1102302937 


32.1403173 


10.1088117 


1034 


1069156 


1105507304 


32.1558704 


10.1120726 


1035 


1071225 


1108717875 


32.1714159 


10.1153314 


1036 


1073296 


1111934656 


32.1869539 


10.1185882 


1037 


1075369 


1115157653 


32.2024844 


10.1218428 


1038 


1077444 


1118386872 


32.2180074 


10.1250953 


1039 


1079521 


1121622319 


32.2335229 


10.1283457 


1040 


1081600 


1124864000 


32.2490310 


10.1315941 


1041 


1083681 


1128111921 


32.2645316 


10.1348403 


1042 


1085764 


1131366088 


32.2800248 


10.1380845 


1043 


1087849 


1134626507 


32.2955105 


10.1413266 


1044 


1089936 


1137893184 


32.3109888 


10.1445667 


1045 


1092025 


1141166125 


32.3264598 


10.1478047 


1046 


1094116 


1144445336 


32.3419233 


10.1510406 


1047 


1096209 


1147730823 


32.3573794 


10.1542744 


1048 


1098304 


1151022592 


32.3728281 


10.1575062 


1049 


1100401 


1154320649 


32.3882695 


10.1607359 


1050 


1102500 


1157625000 


32.4037035 


10.1639636 



23* 



270 TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 

Note. — Since all numbers are the square roots of their squares, and the 
cube roots of their cubes, it follows that the numbers tabulated are the square 
roots of their respective squares, and the cube roots of their respective cubes. 
Moreover, one-half, one-third, one-fourth, &c, of the square root of any num- 
ber is the square root of one-fourth, one-ninth, one-sixteenth, &c, of that 
number ; thus one-tenth of the square root of 150 is the square root of 150 
H-100= x/U, and one-fourth part of the square root of 125.72, is the square 
root of 125.72 -=-16 = ,77.8575, &c. But the process of extracting the square 
roots of numbers by arithmetic is so simple, and the labor so trifling, that 
but little will be gained, generally, by resorting to the tables, especially when 
the square root of a mixed nnmber is required. 

By the help of the Table, to find the Cube Root of a number that is 
not the exact Cube of any number tabulated, and that is not great- 
er than the Cube of one more than the highest number tabulated, 
n = given number whose cube root is required. 
t = tabular cube nearest the given number. 
b = tabular cube root of t. 
r = cube root of given number, or cube root required. 

(2n + Q b 
r = a . — — nearly. 

2t -j- n J 

Note. — This formula expresses the cube root correct to within about 
1-100 of a unit, under its most unfavorable application, and often affords almost 
strict accuracy, even when the maximum error obtains, if the given number 
be large. When two tabular cubes are equally near the given number, or very 
nearly so, use the greater, by which a closer approximate will be obtained, and 
r will be plus by the amount of the error. 

By help of the Table, to find the Square Root or Cube Root of a 
mixed number, whose integer, and the integer next higher, are tab- 
ulated. 

n = given mixed number whose root is required. 
R — root of the integer next higher than the integer of the 
given number. 
i = integer of the given number. 
s = root of the integer of the given number. 
r = root of the given mixed number, or root required. 
r=(R — .s)(n — i) -j- s, practically correct for ordinary purposes. 
Example. — Required the square root and cube root of 18.54. 
^19 = 4.3588989 V 19 = 2.6684016 



y/18 — 4.2426407 
.1162582 
.54 



4650328 
5812910 



.062779428 
y/18— 4.2426407 
^18.54 = 4.305420128. Ans. 



3 y/18 = 2.6207414 
.0476602 
- .54 



1906408 
2383010 



.025736508 
yi8 — 2.6207414 
yi8.54 = 2.646477908. Ans, 



SECTION V. 

MECHANICAL POWERS, CIRCULAR MOTION, &c. 



The Mechanical Powers are the known elements of machinery. 
They are three in number, with some diversity of application. 
Strictly speaking-, they are not powers, or sources of power ; they 
simply convey applied force, and diffuse or concentrate it. In treat- 
ing of them, the term weight, or resistance, is understood to be the 
force to be overcome, and the term power, the force applied to over- 
come or balance it. It is also to be understood that the deductions or 
conclusions arrived at are theoretically true ; that is, that they are true 
upon the supposition that the whole power employed is expended to 
the end under consideration — that no friction or weight of machinery 



THE LEVER. 
Lemma. — The power multiplied by its distance from the fulcrum 
equals the weight multiplied by its distance from the same point ; and 
as the distance between the power and fulcrum is to the distance 
between the weight and fulcrum, so is the effect to the power. 

Consequently, if we divide the weight by 
the power, we obtain a quotient equal the length 
of the longer arm of the lever, the length of 
the shorter arm being 1. And if we multiply 



(ld/ A ' the weight by its leverage, and divide the prod- 

7 m uct by the power, we obtain a leverage for the 

6 power that will enable it to equipoise the 

— — • ^. weight. And if we multiply the power and 

g " Ja>s ^^ its leverage together, and divide the product by 

~~ --=^_ . t ^ e we jgj lt? we 0D t a i n f or the weight the same 

O result. So, too, if we divide the lever by the 
quotient obtained by dividing the weight by the power, to which quo- 
tient we have added 1, we obtain the relative position of the fulcrum, 
or the distance it must occupy from the opposing force. And, again, 
if we multiply the opposing force by its leverage, and divide the prod- 
uct by the leverage pertaining to the power, we obtain the requisite 
power to counterbalance the resistance. 

Example. — A weight of 1200 lbs., suspended 15 inches from the 
fulcrum, is to be raised by a power of 80 lbs. ; at what distance from 
the fulcrum on the long arm of the lever must the power be applied, to 
accomplish that end? 

80 : 1200 :: 1.25 : 18f feet. Ans. 



272 MECHANICAL POWERS. 

Example. — The lever is 20 feet long, the opposing force 
1200 lbs., and the available force 80 lbs. : at what distance from the 
former force must the fulcrum be placed, that the two forces may 
equipoise each other ? 20 X 80 -^ (1200 -j- 80) — \\ feet ; or 

1200 -f- 80 = 15, and 20 -J- 15+1 = 1£ feet. Ans. 

Example. — The longer arm of the lever is 18| feet, the shorter 
arm 1{ feet, and the weight to be raised is 1200 lbs. ; what power 
fnust be applied to raise it? 

18.75 : 1.25 :: 1200 = 80 lbs. Ans. 

Example. — A man, with a lever 5 feet in length, raised a weight 
of 2500 lbs. suspended across the lever 9 inches from the further end, 
which rested on a support ; what force did the man exert ? 
5 : .75 :: 2500 = 375 lbs. Ans. 

Example. — A beam, 20 feet in length, supported at both ends and 
not elsewhere, bears a weight of 6000 lbs. placed 6 feet from one 
end ; what is the pressure on each support 1 

20 : 14 :: 6000 : 4200 lbs. on the support nearest the weight. > . 
20 : 6 :: 6000 : 1800 lbs. on the support furthest from the w't. ^ JLns ' 

WHEEL AND AXLE. 

The wheel and axle is a revolving lever. It 
partakes, in all respects, of the same principles 
as the preceding. The radius of the wheel is 
the longer arm of the lever, and the radius of 
the axle, the shorter. The fulcrum is the point 
of impact between them — at the circumference 
of the axle. 

Example. — The radius of the wheel is 2.^ 

feet, the radius of the axle is 9 inches, and the weight to be raised is 

500 lbs. ; the weight is attached to a rope wound round the axle ; 

what power must be applied to the periphery of the wheel to raise it ^ 

2.5 : .75 :: 500 : 150 lbs. Ans. 

Example. — The diameter of the wheel is 5 feet, the diameter of 
the axle or barrel, 1£ feet, and the power is 150 lbs. ; what weight 
may be raised ? 

1.5 : 5 :: 150 : 500 lbs. Ans. 

Example. — The power is 150 lbs., the resistance is 500 lbs., and 
the barrel has 9 inches radius ; required the diameter of the wheel 
that w r ill enable the power to equipoise the weight. 
150 : 500 :: 9 : 30 in. radius, and 30 X 2 =60 -r- 12 = 5 feet. Ans. 

Example. — The length of the winch (crank) of a crane is 15 




MECHANICAL POWERS. 



273 



inches, the radius of the barrel around which the lifting chain coils is 
3 inches, the pinion has 8 teeth, and the wheel 68 ; required the 
weight that a force of 30 lbs. applied to the winch will raise. 

68 -7- 8 = 8^ (8^ to 1) velocity of pinion to wheel, and 
15 X 8.5 -^- 3 = 42.5 lbs. exertive force, or force to 1 of applied 
power — gained at the expense of space, and 
42.5 lbs. X 30 lbs. (applied power) = 1275 lbs. effective power. Ans. 

Example. — The exertive force, or effect to power, of a crane, is 
to be as 42^ to 1, the radius of the wheel to that of the pinion as 8£ 
to 1, and the throw of the winch — its length — the radius of the cir- 
cle which it describes — is to be 1^ feet ; what must be the diameter 
of the barrel? 

8.5 X 1.25 = 10.625 -f- 42.5 = .25 X 2 = .5 ft. or 6 in. Ans. 

Note. — By additional wheels and pinions, as in the system of pulleys or block and 
tackle, which see, the exertive force of a crane may be increased to almost any conceiva- 
ble extent ; but always, as with the block and tackle, and as shown in the above exa.n- 
ple, at a relative expense to space. 

THE PULLEY. 




A single pulley, fixed and turning on its own axis, affords no me- 
chanical advantage. ^It serves but to change the direction of the 
power. 

In the common system of pulleys, or block and tackle, the advan- 
tage is as the number of ropes engaged in supporting the lower or 
rising block, to 1 of applied force. 

Rule. — 1. Divide the given weight by the number of cords lead- 
ing to, from, or attached to, the lower block, and the quotient is the 
requisite power to produce an equilibrium. 

Rule. — 2. Multiply the given power by the number of cords lead- 
ing to, from, or attached to the lower block, and the product is the 
weight that may be raised. 

Rule. — 3. Divide the weight to be raised by the power to be 
applied, and the quotient is the requisite number of cords that must 
connect with the lower block. 

Example. — The lower, running, or rising block has 5 sheaves or 



274 MECHANICAL POWERS. 

pulleys ; the fixed or stationary has 4 ; and the weight to be raised 
is 2250 lbs. What force must be applied to raise it ? Necessarily 
the end of the rope is attached to the lower block, therefore 9 ropes 
are attached to or connected with it ; hence — 

2250 -£-9 — 250 lbs. ■ Ans. 

Note. — In the Spanish burton, having two movable pulleys and two separate ropes, 
the effect is to the power as 5 to 1. In a system of 4 movable pulleys and 4 separate 
ropes, it is as 16 to 1. And in a system having 4 movable and 4 fixed pulleys, and 4 sep- 
arate ropes, it is as 81 to 1. 

INCLINED PLANE. 
Lemma. — The product of the length of the plane and power is 
equal to the product of the height of the plane and weight. 

The velocity, therefore, or force, or momentum, with which a body 
descends an inclined plane, impelled by its own gravity, is to that 
with which the same body would descend perpendicularly through 
space, as the height of the plane to its length, or as the size of its 
angle of inclination to radius. And the space the body describes upon 
the plane, in any given time, compared with that which it would de- 
scribe falling freely, in the same time, is as its velocity upon the plane 
to that of perpendicular descent. And, the spaces being the same, 
the times will be inversely in that proportion. 

The deductions, therefore, are — 

1. That the product of the weight and height of plane, divided by 
the length of plane, gives the requisite power to sustain or balance 
the weight. 

2. That the product of the power and length of plane, divided by 
the height of plane, (which reverses the former process,) gives the 
weight or resistance that the power will overcome. 

3. That — ,(the times being equal) — the velocity attained, or 
force acquired, or space described, by a body falling freely from rest, 
multiplied by the height of plane, affords a product which, divided by 
the length of plane, gives the velocity attained, or force acquired, or 
space described, by a body moving down the plane, impelled by its 
own gravity. 

4. That the product of the weight and base of plane, divided by 
the length of plane, gives the pressure on the plane. 

To find the base of the plane. 
Rule. — From the square of the length of the plane, subtract the 
square of the height, and the square root of the difference is the base. 

To find the height of the plane. 
Rule. — From the square of the length of the plane subtract the 
square of the base, and the square root of the difference is the height, 



MECHANICAL POWERS. 



275 




To find the length of the plane. 
Rule. — Add the square of the base and the square of the height 
together, and find the square root of the sum, which will be the 
length sought. 

WEDGE. 

The wedge is a double inclined plane. Its principles are the same, 
and they are wholly covered by the preceding. 

The power multiplied by the length of a side, equals 
the resistance multiplied by half the breadth of the head. 

When, therefore, both sides of the substance to be 
cleft are movable, the product of the resistance and half 
the breadth of the head, divided by the length of the side 
of the wedge, gives the requisite force to be applied. 
And when only one side of the substance is movable, the 
product of the resistance and breadth of the head, divided 
by the length of a side of the wedge, gives the power 
required. 

SCREW. 

If we take the figure of an inclined plane — a 
right-angled triangle say, cut from paper — and 
unite the extremities of the base, we have the 
figure of the screw, m principle ; and the prin- 
ciple of the screw is that of an inclined plane 
curved to a cylinder ; and the screw is not a me- 
chanical power, any more than the wedge, or 
wheel and axle. It is the plane that is an element 
of machinery, and not the curve or the cylinder 
around which the plane is placed. And it appears 
that the screw, the inclined plane, and the right- 
angled triangle, are mathematically the same. 
Thus, if we would find the length of the thread of a screw by the 
circumference and pitch, we are to find it as we would find the 
length of the inclined plane by the base and height, or the hypote- ■ 
nuse of a triangle by the base and perpendicular, and so, in like 
manner, for the other lines of the figure. 

The pitch of the screw or rise of the thread in a revolution corre- 
sponds to the height of the plane or perpendicular of the triangle. 
The circumference of the screw corresponds to the base of the plane 
or base of the triangle. And the length of the thread making one 
revolution around the cylinder — the working circumference of the 
screw — corresponds to the length of the plane or hypotenuse of the 
triangle. The mechanical advantage of a screw is as the length of 
the plane to the size of its angle of inclination. 




276 MECHANICAL POWERS. 

The ordinary screw, therefore, — the piece of mechanism, — is this 
plane repeated along a cylinder a greater or less number of times, 
whereby the spiral thread alluded to, now the sectional thread of 
the instrument, becomes the continuous thread, helix or spiral, that 
extends, by construction, at a uniform angle to the cylinder's axis, 
throughout its length. And it is to be borne in mind, always, that 
the pitch of a screw is the distance, parallel to the axis, between 
any two consecutive windings of the same apparent continuous 
thread, measured on its face, from centre to centre. 

If a screw have more than one apparent continuous thread, there- 
fore, — and screws are often constructed with two, and sometimes 
more — the pitch of such screw is still the same as it would be were 
but one such thread employed, or as it would be were all but one 
removed. Twelve rafters to the side of the roof of a building can 
sustain more pressure than four, and in the ratio of three to one, but 
the fitch of such roof is neither greater nor less in consequence of 
the number of rafters employed. 

The screw multiplies the extent of the action of the inclined plane, 
it will be perceived, as many times as the plane is repeated along 
its cylinder, whereby great advantage of mechanical application is 
obtained ; but the mechanical advantage of the screw, it is apparent 
is still with the plane, and not in the number of times the plane is 
used. The power is no more with the latter than it is with the 
curve of the plane (which is another advantage of mechanical appli 
cation) instead of being in the plane itself. 

And there is still another advantage pertaining to the screw, or to 
this mode of employing the inclined plane, viz., that the length of 
the plane or working circumference of the screw may be increased, 
in effect, to almost or quite any desired extent, by the employment 
of a simple bar or rod to turn the screw ; whereby, the power and 
pitch remaining the same, the mechanical advantage is enhanced as 
much as the working circumference is increased ; that is, it is made 
greater as much as the circumference of a circle, the bar being the 
radius thereof. 

The power, force, or mechanical advantage of a screw, as we have 
said, is that of the inclined plane employed, and is as its length to 
the size of its angle of inclination, or it is as the working circumfer- 
ence of the screw to the pitch. 
If we let P represent power, 

L " length of lever, 

/ " length of inclined plane, 

W " weight or pressure, 

p " pitch of screw or angle of inclination, 

r " radius of screw, 

C " circumference described by power, 

x " effect of power at circumference of screw. 



MECHANICAL POWERS. 277 



Then we have — 
l-.p :: W :P 
I : W : : p : P 

P : W ::p : I 



W :/::P :p 
r :L :: P ix 
P :a? ::r : L 



L : r : : a? : P. 
P : W ::p :C 
C : P :: W :P 



Example. — The circumference of a screw is 12 inches, its pitch 
1{ inches, and the power is 30 lbs. ; what weight may be raised I 

yv/(12 J -\- 1^ ) = 12.065, working circumference of screw, and 
1.25 : 12.065 : : 30 : 289.56 lbs. Arts. 

And if a bar 14 feet in length (rectilinear distance from the 
point on the bar at which the power is applied to the circumference 
of the screw) be employed to turn the screw, the power remaining 
the same, what weight may be raised? 

14 X 12 X 2 X 3.1416 = 1056 inches, circumference of circle, 
bar as radius ; and 

1056 -f- 12.065 = 1068.065 inches, circumference described by 
power ; and 

1.25 : 1068.065 : : 30 : 25633 lbs. Ans. 

Or, 12.065 : 289.56 : : 1068.065 : 25633 lbs. Ans. 

The foregoing exhibits the method of finding the strict theoretical 
force or mechanical advantage of the screw. But for most practical 
purposes, more especially if we take into account the fact that the 
actual force in consequence of the friction is only about two-thirds 
that of the theoretical, the rectilinear distance from the point on the 
bar at which the power is applied, to the centre of the screw, may 
be taken as the radius of the circumference described by the power, 
or as the effective or working circumference of the screw, instead of 
the true working circumference as found above ; and this is more espe- 
cially true if the pitch of the screw be but slight or inconsiderable. 
Thus, in the aforementioned screw of 12 inches circumference aiad 
1| inches pitch, the difference between the actual circumference and 
the working is only .065 of an inch, which, when the effect is of some 
magnitude, is of no particular account. The example next below is 
illustrative, and is given in proof of this position. 

Example. — The pitch of a screw is li inches, the power 30 lbs., 
and the rectilinear distance from the centre of the screw to the point 
on the bar at which the power is applied is 169.91 inches ; required 
the weight that may be raised, supposing this rectilinear distance to 
be the radius of the circle described by the power. 
24 



278 MECHANICAL POWERS. 

169.91 X 2 X 3.1416 = 1067.57 inches, circumference by assumed 
radius ; and 

1.25 : 1067.57 : : 30 : 25621 lbs. Ans. And showing an error 
of only 12 lbs. in 25633 lbs., consequent upon having employed the 
assumed radius instead of the real, and that, too, under a pitch so 
unfavorable as the one supposed. 

"When a hollow screw revolves upon one of less diameter and 
pitch, the effect is the same as that of a single screw whose pitch is 
equal to the difference of the pitches of the two screws. 

Thus, if a hollow screw of ^ of an inch pitch revolves upon one 
of -| of an inch pitch, the power to the weight is as -£ ^ £ = ^V 5 
that is, the power being 1, the weight will be 24. 

A screw of this description and with these pitches, therefore, if 
turned with a bar 6 inches in length, (distance from the power to 
the centre of the screw,) will, in order to produce an equilibrium, 
require a power to the weight as 1 to 24 X 2 X 3.1416 X 6 == 905. 
In a complex machine, composed of the screw and wheel and axle, 
the relations of the weight and power are as under : — 
Let R represent radius of wheel. 
r " radius of axle. 
p " pitch of screw. 
C " circumference described by power. 
x " effect of power on the wheel. 
Then — 



PXCXR^WXPX*- 
P: W::pxr:CX R 
2>Xr:CXR::P:V7 



pxc = *x;> 

WX r = RX* 
PXCX*XR = *XWxrXi> 

And, if intermediate movers are inserted, (wheels and pinions, or 
drums and pulleys,) the same principles still apply. 

Example. — The length of the crank (lever) which turns an end- 
less screw, is 24 inches, and the pitch of the screw is | of an inch; 
it turns a wheel of 30 inches radius, which turns a pinion of 7 inches 
radius, which turns a wheel of 22 inches radius, which turns an axle, 
around which the lifting chain winds, of three inches radius ; what 
weight will a power of 50 lbs. applied to the crank raise? 
p r r C RRPW 

.75 X 7 X 3 : 150.8 X 30 X 22 :: 50 : 315962 lbs. Ans. 

Note — It is clear, we may substitute the diameters of the wheels and pinions for 
their radii, if we prefer ; or we may work by the number of teeth in each, in which latter 
case, the circumference of the axle, in the foregoing, would come in to be employed. 

If the screw, acting upon the periphery of the wheel, have more 
than one thread, the real, or obvious pitch, spoken of above, must be 
taken, increased an equal number of times. 



LATERAL OR TRANSVERSE STRENGTH OF BODIES. 



279 



LATERAL OR TRANSVERSE STRENGTH OF BODIES. 

The transverse strength of a body is its power to resist force or 
weight acting upon it in a direction perpendicular to its length. 

Against each particular denomination of material in the following 
table is placed the weight (mean of various experiments) required to 
break a solid, uniform bar, One Foot in Length and One Inch Square, 
of that material, the bar being fixed at one end and the weight sus- 
pended from the other, the action of the weight direct with the bar's 
sides. 

The Woods of American growth, and seasoned. 





Breaking 


Greatest 




Breaking 


Greatest 


Materials. 


Weight in 


Deflection 


Materials. 


Weight in 


Deflection 




lbs. 


in Inches. 




lbs. 


in Inches. 


Hickory, 


270 


8. 


Pitch Pine, 


225 




White Ash, 


234 


2.5 


Yellow Pine, 


150 


1.70 


White Oak, 


220 


9. 


White Pine, 


138 


1.40 


Chestnut, 


170 


1.7 


Cast Iron, 


684 


0.63 


Elm, 


142 




Wrought Iron, 


1012 





About 600 lbs. suspended from the end of a square bar of wrought 
iron, of dimensions and fixed as supposed, causes the bar to deflect 
about one inch, at which it takes a permanent set, or bend. 

As the weight written against any particular denomination of ma- 
terial, in the foregoing table, is the weight required to break that 
material, under the length, lateral figure, and condition supposed, it 
follows that the same weight may be taken as the constant or co- 
efficient in determining the weight required to break the same denom- 
ination of material, under different lengths, lateral figures, relative 
conditions, &c. 

C = tabular constant, or initial weight, above. 
I = length of bar or beam in feet. 
b ■=■ breadth of rectangular bar in inches. 
v = vertical dimensions, or depth of rectangular bar in inches. 
W = breaking weight, or ultimate transverse strength of bar under 
investigation. 

1 . When the bar is fixed at one end, and the weight suspended from 
the other ; the weight of the bar not being taken into account. 

j : v' x b :: C : W. 

2. When the beam is supported (not fixed) at both ends, and the, 
weight in the middle. 

I : bv 2 :: 4C : W. 

3. Fixed at both ends, and the load in the middle. 

l:bv*::6C : W. 



280 LATERAL OR TRANSVERSE STRENGTH OF BODIES. 

4. Fixed at one end, and the load distributed uniformly over its 
whole length. 

I : v 2 b :: 2C : W. 

5. Supported at both ends, and load distributed uniformly over whole 
length. 

I : v 2 b :: 8C : W. 

6. Supported at both ends, and load at the distance mfrom one end. 

mX(l-m) : v 2 bl :: C * W. 

Or, 2(1 — m) X 2(1 — n) -7- / = I 1 = effective length, and 

Z 1 :> a :: 4C : W. 

Note. — bv 2 in a square beam = a side of the square beam cubed. 

Example. — A beam of white Oak, 3 feet in length, 4 inches deep, 
and 2 inches in breadth, is fixed at one end ; what weight is required 
to break the beam, the weight being suspended from the other end? 

3 : 4 2 X 2 :: 220 :: 2346| lbs. Ans. 

Example. — The same beam, same manner of support, &c, as the 
foregoing, but the greater cross-section of the beam placed horizon- 
tally ; what weight is required to break it ? 

3 :2 2 X4 ::220 : 1173 J lbs. Ans. 

Example. — A beam of cast iron, 8 feet in length and 6 inches 
square, is supported at both ends ; what is its ultimate transverse 
strength, it being loaded in the middle ? 

8 : 6 2 X 6, i. e., 8 : 6 3 :: 684 X 4 : 73872 lbs. Ans. 

A beam fixed at both ends, other things being equal, will bear one 
half more than when merely supported at both ends. 

The length of a beam supported at each end, or by two supports, is 
the distance from one support to the other. 

Round beams, supported in the middle and loaded at each end, 
have the same sustaining power as when supported at each end and 
loaded in the middle ; and the same is true for rectangular beams, the 
action of the load, in both cases, being, in the same manner, direct 
with the beam's central plane. 

An equilateral triangular beam, supported in the middle, and 
loaded at each end, has the same sustaining power as when supported 
at each end and loaded in the middle, if the beam be inverted. 

When a beam is partly loaded at any given locality of its length, to 
find what weight, acting upon it at any other given locality, must be 
added to break it. 

Let E = breaking weight at the locality of the given partial load. 
F = breaking weight at the locality of the required partial load. 



LATERAL OR TRANSVERSE STRENGTH OE BODIES. 281 

Let a = given partial load. 

x = required partial load ; then 

E : F :: E — a : x. 

With regard to required depths, breadths, tffc. 

Let S = effective coefficient in all cases ; that is, = C, or any mul- 
tiple of C, demanded by the conditions, as set forth in Prob. 1, 2, 3, 
< 5, or 6, foregoing; then 



I : v°-b :: S : W. 
W : S :: bv 2 : /. 
bv 2 : W :: I : S. 



S : I :: W : bv 2 . 
Sv 2 :W ::l:b. 

Sb : W :: I : v 2 . 



/WZvA vb 1 v 1 

V I -pjTT- ) = v, and — = b; — being the ratio fixed upon for the 
\ So 1 / v l b 1 

depth to the breadth. 

Example. — What must be the depth of a pitch-pine beam, resting 
on two supporters 20 feet apart, that its ultimate transverse strength 
may be 24000 pounds, the beam being 4 inches in breadth, and the 
load resting uniformly along its whole length ? 

By referring to the table of initial weights, we find the prime co- 
efficient for pitch pine to be 225 lbs., and by problem 5 we find that 
8 times that quantity, or 1800 lbs., is the effective coefficient for the 
case in hand. Hence 

1800 X 4 : 24000 :: 20 : V66| = 8.165 inches. Arts. 

Comparative transverse strength of figures, or of beams, <Sfc, of differ- 
ent figures and positions ; both members of the couplet supposed to be 
of the same material, same length, in the same manner loaded, and in 
the same way supported. 

D, a square beam, the weight acting direct with the sides. 

0, a square beam, the weight acting direct with the diagonal. 

A, an equilateral triangular beam, the weight acting direct with 
the perpendicular, and tending to convex a side. 

V, an equilateral triangular beam, the weight acting direct with 
the perpendicular, and tending to convex an angle. 

O, a round beam, or solid cylinder. 

a = side of square beam ; s — side of equilateral triangular 
beam; h = perpendicular of triangle; d = diameter of round 
beam ; t = diagonal of square ; A = area of transverse section. 

The maximum transverse strength of an equilateral triangular 
24* 



282 



LATERAL OR TRANSVERSE STRENGTH OP BODIES. 



beam is obtained when the load tends to convex a side directly, 
and the minimum when it tends to convex an angle directly. 

The maximum transverse strength of a square beam is afforded 
when the action of the load is direct with the sides, and the mini- 
mum when it is direct with the diagonal of the cross section. 



In an equilateral triangle, — = s 2 ; and s 2 
therefore A = %hs. )/% 

In a square, \J2A = t, the diagonal. 



|s 2 or 



SA 
y/3 



= h i 



In a circle, 



AA 






Comparative Transverse Strengths, Sectional Areas equal. 



Formulas of the absolute transverse strengths 
of the respective members. 



Relative transverse strengths of the 
couplets, reduced to the unit of com- 
parison. 



_□_ a 3 ._ As! A 


1 1 


O 2a\t—a) 2A(\/2a — \fA) 2(y/2 — 1) .828427 


□ a 3 As/ A 1 1 


O j%7rd' d 3(A\/A-+-\/tt) 3-^-2^ .846284 


A —W— ISA^A -i- 8^3 _ 15 -f- 8^3 _ 1.424692 


□ u? A\/A 1 1 


□ __ a 3 . A)/ A 1 1 


V T yi 2 s 15A\/A~8\/S^S 15-f-8v/3f3 .822546 


A _ |A 3 _ _ 15.4^-^-8^3 _ \/3 _ 1.732051 
V 3-A-s 15A\/A^r8i/Sf3 ~T~ 1 


A_ W __15A\/A^-S^3 .__ 5v/7r-f-4^3 _ 1.683467 
O &*d* ftAtfA + M 1 1 


O _ tW 3 — K^V/^ -r- V^) _ 4v/3v / 3-f-5y/7r _ 1.028859 


V J^tfs 15A^A-^-8^S^S 1 1 


0___2a 2 (t — a)_2A(\/2A — \/A) 


v T \ h 2 s 15A\/A-±-8\/3p 


16^3^3(^/2 — 1)-^- 15 1.0071495 


1 1 

Comparative Transverse Strengths, side of Square Bar, side of 


Equilateral Triangular Bar, and diameter of Round Bar, equal. 


□ 16 _ 1.697653 Tnv _ 1 
O 3tt 1 .589049 


O _ 47T _ 1.45104 Iny 1_ 

A 5V3 1 .689161 


□ 64 2.463364 „ 1 
A" ~~ 15*/3 ~ 1 ~~ .405949 


O 4tt 2.513274 „ 1 
V "~ 5 " 1 "".397887 



LATERAL OR TRANSVERSE STRENGTH OF BODIES. 283 

Example. — Required the transverse strength, or breaking 
weight, of a solid cylinder of cast iron 3 inches in diameter and 4 
feet in length, one end being fixed, and the weight suspended from 
the other. 

^_ CVW 3 _ 684 X 3 y 3.1416 X 3 3 _ 9yifl fij>7 ^ ^ 
I 4X 16 

Cf.73 

Also, W = _, S being C X .589049, the relative strength of 

a round beam to that of a square beam, the diameter of the one 
being equal to a side of the other, page 282. 

Example. — What is the transverse strength of an equilateral 
triangular beam of white oak, 1 2 feet in length and 6 inches to a 
side, the beam resting on two supports, angle up, and the load sus- 
pended from the middle ? 

W = ***£ = 5X140.296X4X220 = MQM ^ ^ 

81 8 X 12 

Also, W — ^, S being C X .405949, the relative strength of 

an equilateral triangular beam, angle up, to that of a square beam 
of the same width of side. 

HOLLOW CYLINDERS. 

The lateral strength of a hollow cylinder, within certain limits 
of expansion, is to that of a solid cylinder of the same material, 
quantity of matter, and length, other things being equal, as the 
difference of the cubes of the diameters to the cube of the square 
root of the difference of the squares of the diameters, nearly. 

D = diameter of hollow cylinder. 

o = diameter of bore, or diameter of interior. 

m = diameter of the hollow cylinder if converted into a solid 
cylinder without changing its length. 

d = diameter of a solid cylinder, other things being equal, that 
has a transverse strength equal to that of the hollow cylinder. 

W, S, I, as in the preceding. 

& = m?-{- o 2 ; d s = D 3 — o 3 ; m 2 = D 2 — o 2 ; (? = & — m 2 . 

Example. — What is the transverse strength of a hollow cylin- 
der of cast iron ; the diameter being 8 inches, interior diameter 6 
inches, and length 12 feet; it being supported at both ends, and 
the load suspended from the middle ? 

4S(I? *) = 402.91(512 - 216)4 = ^ ^ ^ 
I 12 

Example. — What would be the diameter of the above de- 
scribed cylinder if it were converted into a solid cylinder of the 
same length ? 

\/(8 2 — 6 2 ) == 5.2915 inches. Arts. 



284 



LATERAL OR TRANSVERSE STRENGTH OP BODIES. 



By which it appears that the lateral strength of a solid cylinder 
5.29 inches in diameter is doubled if expanded into a hollow cylin- 

der of the same length and 8 inches in diameter ; or — - — — — 

296 LV( 8 " — 6 ")J 

— 148.162 = 1 - 998 ' 

Note.— In practice, with a view to safety, a material should not be relied on 
as having a permanent lateral strength exceeding one-third its ultimate lateral 
strength. 

RESULTS OF EXPERIMENTS BY MAJOR WADE. 

Square Bars, sectional area 1 inch. 



Materials. 


Specific 
gravity. 


Tensile 
force. 


Transverse 
strength. 


Crushing 
weight. 


Hardness. 


Cast steel 

Wrought iron .... 

Cast iron 

Bronze 


7.727 

8.953 

7.704 

7.858 

6.9 

7.4 

7.978 

8.953 


128000 
38027 
74592 
9000 
45970 
17698 
56786 


1916 
542 

416 
958 


198944 
391985 

40000 
127720 

84529 
174120 


10.45 

12.14 

4.57 

33.51 

4.57 

5.94 



TORTIONAL STRENGTH. 

Solid Cylinders, length 1 foot, diameter 1 inch. 



Materials. 


Specific 
gravity. 


At J 
degree. 


Ultimate. 


Specific 

gravity. 


At i 
degree. 


Ultimate. 


Cast steel 

Wrought iron 

Cast iron 

Bronze 


7.727 
7.704 
6.9 
7.978 


970 

1006 

620 


5511 

1296 
1660 
1687 


7.858 

7.4 

8.953 


1320 

2500 

833 


1836 
3060 
1020 



The elasticity of wrought iron is equal to a tensile force of 
about 21000 lbs. per square inch of transverse section, and that 
of cast iron is balanced by a tensile force of about 5000 lbs. per 
square inch of cross section. 

To find the ultimate resistance to internal pressure, or bursting, of 
hollow tubes, cylinders, steam boilers, 8fc. 

C — cohesive or tensile force per square inch of the material, 
foregoing table, or table page 74. 

/ = length of the tube, or shell, in inches. 

d = diameter of the tube, or vessel, in inches. 

t = thickness of the plate, or side of the tube, in inches. 

W = ultimate resistance, or bursting pressure, in lbs., per square 
inch. 



LATERAL OR TRANSVERSE STRENGTH OF BODIES. 285 

RESULTS OF EXPERIMENTS ON WELDED WROUGHT IRON 
TUBES BY W. FAIRBAIRN. 

Ends secured to head-plates. 



Resistance to external pressure. 



Resistance to internal pressure. 



1 


d t 


W 


I 


d 


t 


w 


12 . 


... 6 043 .. 


.. 475 


30 .. 


.. 6 .. 


.. .043 ... 


.. 65. 


24 . 


. .. 6 043 .. 


.. 235 


59 .. 


.. 6 .. 


.. .043 .. 


.. 32. 


30 . 


. .. 6 043 .. 


.. 230 


30 .. 


.. 12 .. 


. . .043 .. 


.. 22. 


60 . 


...12 043 .. 


.. 110 


60 .. 


. . 12 .. 


. . .043 .. 


.. 12.5 



Although these experiments, at first sight, seem to warrant the 
conclusion that the length of a tube, or shell, is properly an ele- 
ment to be employed in computing its resistance to internal pres- 
sure, or bursting, yet the length, within ordinary practical limits, 
can only, in a very slight degree, affect the tensile force of the 
material ; and this position is very nearly sustained by the last 
three of the experiments alluded to. The length, for all ordinary 
purposes of calculation, may be rejected. 

Example. — The diameter of the wrought-iron shell of a steam 
boiler is 36 inches, and the thickness of the plate ^ inch ; what is 
the ultimate resistance of the shell per square inch to internal pres- 
sure, or bursting ? 

A(Jt C y 2t 

W= , or by the common rule W= — ^— then 

y/(-n-d 2 ) d 

55000X4X.25 =862 lb St Ans. 
y/(3.1416 X 36 2 ) 

Note. — This rule, it should be borne in mind, gives the ultimate resistance to 
internal pressure of the plate or material (supposing the initial force C to be 
correctly taken), and makes no allowance for weakness due to the manner of 
building. In the case of a single-riveted tube, or steam-boiler, it is customary 
to take one-half the ultimate resistance of the plate as the ultimate strength of 
the structure; and a double-riveted is one-third stronger than a single-riveted. 
The cohesive or tensile strength of wrought-iron boiler-plates ranges from 
42000 to 62000 lbs. per square inch. These remarks are also applicable in the 
following proposition. 

To find the ultimate resistance to external pressure, or collapsing, of 
tubes, Jlues, hollow cylinders, fyc. 

W = — — ^— nearly, C being the crushing force per square 
Ttld 
inch of the material, foregoing table, or table page 289. 

Example. — What is the resistance to collapsing of a wrought- 
iron tube, the thickness being ^ inch, diameter 7^- inches, and 
length 100 inches, assuming the initial crushing force of the ma- 
terial to be 83500 lbs. ? 

8 3500 X ^83500 X -25 2 = 7680312 X .25_ 2 __ 64Q lbg ^ 
3.1416 X 7.5 X 100 100 X 7.5 ' 



DEFLECTION OF BEAMS, SHAFTS, &c. 

w = weight with which the beam is loaded, in pounds. 
a = tabular or initial deflection, page 279. 
A = actual deflection in inches. 
W, C, I, b, and v, as before. 

1. When the beam is fixed at one end, and loaded at the other ; the 
weight of the beam not being taken into account. 

wl 3 wPa 

Deflection varies as -7-5 ; an( * Tan = A. 
bv 3 ' Cbv 3 

Ratio of Load to Breaking Weight, or Deflection to Greatest De- 
wl Cbv 2 ■ Cbv 2 „ , , 

flection, as pTrv ~~r~ == " • — 7 w == Difference of load and 

breaking weight. 

2. Fixed at one end, and load distributed uniformly over whole 

length. 

wl s Swl 3 a 

Deflection varies as 7-3 ; and u/1 , , = A. 
bv 3 ' SCbv 3 

wl 
Ratio of Load to Breaking Weight, &c, as *, ,, a . 

3. Supported at both ends, and loaded in the middle. 

^ „ . . w>Z 3 , wal 3 

Deflection varies as t-j ; and oori , 8 = A. 

bv 3 ' 32CZw d 

t#Z 
Ratio of Load to Breaking Weight, &c, as .p, 2 . 

4. Supported at both ends, and loaded uniformly along the whole 
length. 

wl s 5 wal 3 

Deflection varies as 7-r, ; and 5- X 0^7 * = A. 
fry 3 ' 8 ^ 32C£y 3 

wZ 
Ratio of Load to Breaking Weight, &c., as ^p- 

The deflection of a round beam is to that of a square beam, the 
diameter of the former being equal to a side of the latter, and other 
things equal, as the transverse area of the square beam to | the trans- 
verse area of the round beam ; as ~ » 7^4 5 as ~T~' near ty* 

The deflection of a hollow cylinder is to that of a solid cylinder of 
the same material, quantity of matter and length, other things being 
equal, as the diameter of the latter to the greater diameter of the 
former ; that is, their deflections vary inversely as their strengths ; 
and the deflections of hollow cylinders, one with another, other 
things being equal, are inversely as their exterior diameters nearly. 



RESISTANCE TO TORSION. 

The following table shows the weight or force in pounds (mean of 
experiments) , required to twist asunder bars One Inch Square and 
One Inch in Diameter of the materials named, the Weight acting 
upon the bar at One Inch from the bar's axis. 



Materials. 


Sq. Bar. 

lbs. 


Rd. Bar. 

lbs. 


Cast Steel, 


28560 


17620 


Blistered Steel, 


24380 


15040 


American Wrought Iron, 


14780 


9120 


Swedish Wrought Iron, 


13870 


8560 


Cast Iron, 


13780 


8500 


Yellow Brass, 


6860 


4230 


Cast Copper, 


6280 


3875 


Oak, 


6160 


3800 


Fir, 


6690 


4130 



American Wrought Iron 
twists and takes a per- 
manent set — square bar 
with 9640 lbs., round bar 
with 5950 lbs. applied. 
Swedish Wrought Iron 
twists and takes a per- 
manent set — square bar 
with 9790 lbs., round, 
with 6040 lbs. applied. 



C = tabular weight above , special for the case. 

W = breaking weight in pounds. 

r = leverage of applied force or W's radius of action, in inches, 
— (the perpendicular distance in inches from the axis of the shaft to 
the point on the lever or crank, where the motive power is directly 
applied.) 

d = side of square shaft or diameter of round shaft, in inches. 



W = 



Cd s . 
r 

Cd?. 

W 



c - d* 

Wr 



Example. — Required the weight or force necessary to twist asun- 
der a bar of square rolled iron, 2 inches to the side, the force acting 
upon the bar through a lever or crank thirty inches in length. 

14780 X 8 -T- 30 = 3941 lbs. Ans. 

Example. — What must be the diameter of a cast-iron cylinder 



288 RESISTANCE TO TORSION. 

in order that it may have a torsional strength equal to 20000 pounds 
acting upon it through a leverage of 13 feet ? 

20000 X 156 + 8500 = #367.06 = 7.16 inches. Ans. 

Wr 
For Practical Purposes, with a view to safety, -r-77 = d 3 . 

The above cylinder, therefore, for practical purposes, other things 
remaining unchanged, should have a diameter of 

20000 X 156 X 3 -J- 8500 = #1101.18 = 10.32 inches. 

To find the number of degrees torsion that a given weight, with a 
given radius of action, will occasion in a solid cast-iron shaft of given 
length and diameter. 

wlr wlr 

pnr, , 4 = N ; fi „„ „ =sb d i , &c. ; w being the weight or applied 

force in pounds ; I, the length of the shaft in feet ; r, w's radius of 
action in inches ; N, the number of degrees torsion. 

Example. — What must be the diameter of a solid cast-iron shaft, 
in order that if 1200 pounds force be applied to it, through a lever- 
age of 22 inches, to set it in motion, the torsion shall be but 4 de- 
grees in its entire length, its length being 20 feet ? 

# (1200 X 20 X 22 -f- 660 X 4) = 3.76 inches. Ans. 

Note. — Eor Practical Rules applicable to Revolving Shafts, Journals or Gudgeons, see- 

JOCKNALS OF SHAFTS, page 320. 

HOLLOW CYLINDERS. 

The United States Ordinance Manual furnishes the following for- 
mula, deduced by Lieutenant Rodman from the results of experi- 
ments by Major Wade, for finding the torsional strength of hollow 
cylinders, substituting the foregoing symbols to avoid definitions, 
namely, 

D 4 — o i 
W = C X ~j\ — ; JD being the external, and o the internal diam- 
eter of the cylinder ; W, C and r, as before. 

Of course this formula does not admit of resolution so as to find 

Wr D 4 — o 4 .",',.. 

D ; -p- = — =r — =>d 3 ; it is not mathematical, therefore, in 

any degree, to the transverse area of the nfatter in cylinders ; 
/ Wr N 

Basing our calculations upon the results of these experiments, or 
upon the foregoing formula for strength, it may be shown that if we 



RESISTANCE TO LONGITUDINAL COMPRESSION. 



289 



take from the centre of a solid cylinder (by boring or otherwise) , a 
cylinder equal in diameter to £ the diameter of the cylinder from 
which we take it, the hollow cylinder left will retain .987654321 — of 
the torsional strength it possessed before any of its material was ab- 
stracted ; and that if we take, in like manner, from a solid cylinder, 
a cylinder equal in diameter to £ the diameter of the cylinder from 
which we take it, the cylinder thus abstracted from will still retain 
-j- 1 of the torsional strength it possessed when solid ; also, that if we 
take from the centre of a solid cylinder £ its material = txjxtVo" its 
diameter, we shall diminish its torsional strength but 25 per cent. 



RESISTANCE TO LONGITUDINAL COMPRESSION. 

Solid bodies of the same material and length resist longitudinal 
pressure one with another, as are their transverse areas one with 
another, respectively ; but their transverse areas remaining the same, 
their power of resistance is slightly diminished by an increase of 
length ; and when the length of a bodj- exceeds by about four to six 
times the square root of its transverse area (as a general thing for most 
substances) , it cannot be crushed by longitudinal compression ; but, 
being left free, will bend and break transversely. 

The following table shows the force in pounds required to crush 
bars or blocks, in the direction of their lengths, of the materials 
named, the transverse area of each piece being one inch, and the 
length of each not more than three times nor less than once and a 
half the square root of its transverse area. The woods supposed to 
be seasoned, and the length of each bar of wood, in all cases, in the 
direction of its fibres longitudinally. 



Materials. 


Pounds. 


Materials. 


Pounds. 


Ash, Walnut, . 


. 6650 


White Spruce, . 


. 5950 


Elm, Live Oak, 


. 6840 


White Pine, 


. 5780 


Beech, 


. 6960 


Good Red Brick, 


800 


Birch, 


. 7980 


Seneca Sandstone, 


. 10700 


Chestnut, . 


. 5350 


Freestone, . 


3050 


Hickory, . 


. 8925 


Quincy Granite, . 


15300 


White Oak, 


. 6100 


Cast Iron, . 


129000 


Maple, Yellow Pine, 


. 8150 • 


Wrought Iron, 


83500 


Note. — The elasticity of cast iron is 


harely overcome hy a pressu 


re of 5000 


3. to the square inch, and it will hear i 


i compression of -g-§-g hefore c 


rumhling. 


Wrought iron will hear a 
25 


compression 


of -gio- without permanent a 


Iteration. 



290 



RESISTANCE TO LONGITUDINAL COMPRESSION. 



C = tabular weight, foregoing. 

/ = length of column in feet. 

s = side of square column in inches. 

d = diameter of round column in inches. 

b = breadth of rectangular column in inches. 

t = thickness of rectangular column in inches. 

W = reliable or practical resistance. 

C X transverse area = crushing weight. 

For ordinary practical purposes. 

Solid Square Column. 



4 : s 2 :: C : W. 



W :: 4 : C. C : W :: 4 : s 2 . 



Solid Rectangular Column. 

4 : bt :: C : W. ' U : W :: 4 ; C. C : W :: 4 : bt. 

C&:W::4:£. Ct:W::A:b. 

Solid Cylindrical Column. 
5.1 :d 2 :: C : W. C : W :: 5.1 : d\ 

Example. — Required the reliable power of a "White Oak post t<o 
resist longitudinal pressure, the breadth being six inches and the 
thickness 5 inches. 

6100X6X5-r-4 = 457501bs. Ans. 

Example. — Required the reliable or practical resistance to longi ■ 
tudinal compression, of an Elm shaft, 6 inches in diameter. 

6840 X 6 2 -r- 5.1 = 48284 lbs. Ans. 

In keeping with the mass of matter, and the liability of flexure r& 
any length of column. — Practical or reliable resistance. 



Materials. 


Solid Square Column. 


Solid Cylindrical Column. 




17800/s 3 


11125rf 4 


Wrought Iron, 


4s 2 -j-.16/ 2 vv ' 


4d 2 4-.16Z 2 ^ W# 




15300/s 3 


9562d 4 


Cast Iron, 


4iH-l8? "" W ' 


4rf 2 -f.l8/ 2 vv * 




3960/5 3 


2475d 4 


White Oak, 


4s 2 -t-.5Z 2 W * 


4d J L|_.5F — ^ • 



OF CENTRES OF SURFACES kND CENTRES OF GRAVITY. 

SURFACES. 

To find the centre of an equilateral triangle. 
Side X .57735 = distance from the .vertex of each angle. 
Side X .28868 = distance from the middle of each side. 

In any triangle, if a line be drawn from the vertex of either angle 
to the middle of the opposite side, it will pass through the centre 
of the triangle, and the centre will be on that line at § its length 
from the vertex, or \ its length from its junction with the side. It 
is at that point, therefore, where any two such lines cross each 
other. 

To find the centre of a square. 

Side X -5 = distance from the middle of each side. 

Side X .707107 = distance from the vertex of each angle. 

To find the centre of a regular pentagon. 

Side X .688188 = distance from the middle of each side. 
Side X .850644 = distance from the vertex of each angle. 

To find the centre of a regular hexagon. 

Side X .86604 = distance from the middle of each side 
Side X 1 = distance from the vertex of each angle. 

To find the centre of a regular octagon. 
Side X 1.2071 = distance from the middle of each side. 
Side X 1.3065 = distance from the vertex of each angle. 

To find the centre of a rectangle. 

£ V (A 2 -\- a 2 ) , or V (I A) 2 -|- (£ a) 2 = distance from the vertex 
of each angle. 

£ the length of A from the middle of a, and 
£ the length of a from the middle of A, A being one of the 
longer sides, and a one of the shorter. 

To find the centre of a rhombus. 

Side X -5 = distance from the middle of each side. 
From the angles ; on a diagonal at half its length. 

To find the centre of a rhomboid. 
From the middle of A, distant £ a, and from the middle of a. dis- 









292 CENTRES OF SURFACES. 

tant ^ A, to the same point ; A being one of the longer sides, and a 
one of the shorter. From the angles, same as in the rhombus. 

To find the centre of a trapezoid. 

b A + 2a 

k- X a I — = distance from the middle of A, on the line b ; A 

being the longer of the two parallel sides, a the shorter, and b a line 
passing from the middle of A to the middle of a. 

To find the centre of a trapezium. 

Draw a diagonal and find the centre of each of the triangles 
thereby formed, and pass a line from one centre to the other. Then 
draw the other diagonal, and find the centre of each of the triangles 
thereby formed, and connect those centres together by passing a 
line from one to the other. The point at which the lines cross each 
other is the centre of the figure. And this principle is applicable 
to all quadrilaterals. 

In a circle — 

Circumference X .159155 = radius. 

To find the centre of a semicircle. 

Radius 

o oggo = distance on bisecting radius from the middle of the 

chord or centre of the circle. 

/ /radius y\ 

V( Radius 2 -{- ( a - ) ) = distance from both extremities of 

the arc ; n being the ratio of circumference to diameter, that is, 7r 
=-3.1416. 

Radius X .42441 = distance on bisecting radius from the middle 
of the chord. 

Radius -j- 1.737354, or radius X .575588 = distance on bisecting 
radius from the vertex. 

To find the centre of the segment of a circle. 

Chord of the segment 3 

no ks »,^ ~e „„ 4.= distance from the centre of the circle, 

12 X area ot segment ' 

on the versed sine. 

V(£ chord of segment 2 -}- distance of centre of segment from 
middle of chord 2 ) = distance of centre from both extremities of the 
arc. 

Radius of circle — distance from centre of circle = distance from 
Yertex. 



CENTRES OF GRAVITY. 29S 

To find the centre of a sector of a circle. 

| chord of arc X radius of circle . ■ . 

, -r — n = distance from centre of circle, 

length or arc 

on radius perpendicular to the chord, and bisecting the sector. 

V (£ chord of arc 2 -f- (radius — versed sine — distance from 
centre of circle) 2 ) = distance from both extremities of the arc. 

Radius — distance from centre of circle = distance from vertex. 

To find the centre of a parabola. 

On its axis at -§- its length from the middle of the base. 

V (£ base 2 -f- distance from base 2 ) = distance from both extrem- 
ities of the curve. 

Axis — distance from base = distance from vertex. 

In an Ellipse, the centre is at a right section of the figure, as in 
the parallelogram. 

In a Zone, the centre is at a right section of the figure, as in a 
trapezium. 

SOLIDS. 

The centre of gravity of a Cube is its geometrical centre, and the 
same is- true for a Right Prism, Parallelopiped, Cylinder, Sphere, 
Spheroid, Ellipsoid, and any Spindle. It is also true for the Middle 
Frustum of any spindle, the Middle Frustum of a spheroid, the two 
Equal Frustums of a paraboloid, and the two Equal Frustums of a 
cone. 

. In a Cone and Pyramid, the centre of gravity is in the axis, at | 
the length of the axis from the base. 

In a Paraboloid, the centre of gravity is in the axis, at £ the 
length of the axis from the base. 

To find the centre of gravity of a frustum of a cone or frustum of a 
regular pyramid. 

(D -f <f) 2 -f 2 D 2 height 

(D \ dV D d ^ — 4 — ==s distance on * ne ax i s fr° m * ne centre 
of the less base ; D being the diameter of the greater base, and d 
the diameter of the less, in a frustum of a cone ; or, D being a 
side of the greater base, and d a side of the less, in a frustum of a 
regular pyramid. 
25* 



294 CENTRES OF OSCILLATION AND PERCUSSION. 

To find the centre of gravity of a prismoid. 
( VA -f- V«) 2 + 2 A height 

(7a + v«) 2 -vaxv« x_ 4~ = distance on the axi8 

from the less base ; A being the area of the greater base, and a the 
area of the less. And this rule is applicable to the frustum of a 
pyramid, of any number of equal sides, and to the frustum of a 
oone. 

To find the centre of gravity of a frustum of a paraboloid. 

D 2 X2 + rf' height 

— j-w . vi - * — o — = distance on the axis from the less base ; 

D being the diameter of the greater base, and d the diameter of the 
less. 

To find the centre of gravity of a spherical segment. 

3.1416 X height of segment 2 X (radius of sphere — i height of segment)2 

■ — - — - — — — — ■ = dis- 

solid contents of segment 

tance from the centre of the sphere, on the axis of the segment ; the solid contents and 
lines being in the same denomination of measure. 

Radius of sphere — distance from centre of sphere to centre of 
gravity of segment = distance on the axis of the segment from the 
segment's vertex. 

Radius of base of segment 4- height of segment . , , 

Note ; — ; ' = radius of sphere. 

X,01E « height of segment X 2 F 

To find the centre of gravity of a spherical sector. 

(Radius — h versed sine) X 3 .... ,. ^ , ,_ 

i — = distance on bisectmg radius from centre of sphere. 

To find the centre of gravity of a system of bodies. 

• 

— -; — , i „ . " = centre required 5 a, a', a", &c, being the solid 

a -|- a' -\- a", &c. 
contents or weights, and b, b', b", &c, the distances of their respective centres of gravity 
from the given plane 



OF CENTRES OF OSCILLATION AND PERCUSSION. 

The centre of oscillation - applies to bodies fixed at one end and 
vibrating in space ; and the centre of percussion applies to bodies 
revolving around a fixed axis. The two centres, in the same body, 
are at the same point. 



CENTRES OF OSCILLATION AND PERCUSSION. 295 

The centre of oscillation or percussion is that point in a body, 
under one or the other of the supposed motions, in which the whole 
motion, tendency to motion and force in the body, may be supposed 
collected. It is that point, therefore, in the body under motion, 
that would strike any obstacle with the greatest effect, or, if met by 
a staying force, would serve to stop the motion and tendency to mo 
tion, of the whole mass, at the same instant. 

In Pendulums, or rods of uniform diameter and density, having 
one end fixed, and a ball or weight appended to the other : 

Weight of rod X 2 length of rod = momentum of rod. 

Weight of ball X (length of ro<i + radius of ball) = momentum of ball. 

Weight of rod X length of rod 2 
= force of rod. 

Weight of ball X G en g th of rod + radius of ball) 2 = force of ball. 

Force of rod 4- force of ball ,. „ 

— -i = distance from the point of suspension or 

momentum of rod -\- momentum of ball 

axis of motion, to the centre of oscillation or percussion. 

And in a rod of uniform diameter and density, having one end 
fixed, and two or more balls or weights attached : 

Weight of rod X 2 length of rod = momentum of rod. 

Weight of 1st ball X distance from axis of motion = momentum of 1st ball. 

Weight of 2d ball X distance from axis of motion = momentum of 2d ball. 

Weight of lod X length of rod 2 
- = force of rod. 

Weight of 1st ball X distance from axis of motion 2 = force of 1st ball. 
Weight of 2d ball X distance from axis of motion 2 = force of 2d ball. 

Force of rod 4- sum of the forces of the balls 

— — — — - — — - = distance from the axis of motion to 

momentum of rod -f- momenta of the balls 

the centre of oscillation or percussion. 

By another Rule. — Suspend the body freely by a fixed point, and 
cause it to vibrate in small arcs, and note the number of vibrations 
it makes per minute. Then — 

Number of vibrations per minute 2 : 60 2 : : length of pendulum 
that vibrates seconds in the respective locality : distance from point 
of suspension to centre of oscillation. 

Example. — The rod is 10 feet in length, and weighs 21 pounds ; 
the weight of a ball, having its centre at the lower extremity of the 
rod, is 16 pounds, and that of another, affixed to the rod 4 feet from 
the point of suspension, is 9 pounds ; required the centre of oscilla- 
tion in the system. 



296 CENTRES OF OSCILLATION AND PERCUSSION. 

|(21 X 10 2 ) + (9 X 4») + (16 X 10 2 ) * op , • :'■ 

. 21 X 5 + (9X4) + (16 X 10) = 8 ' 12 feet from the 
point of suspension. Ans. 

Example. — If, on suspending the body supposed in the last ex- 
ample by its unloaded end, it is found to vibrate 38 times in a min- 
ute, what is the distance from the point of suspension to the centre 
of oscillation ? 

38 2 : 60 2 : : 3.25846 : 8.12 feet. Ans. 

In a Right Line, Cylinder or Equilateral Rectangular Prism, one 
end being in the axis of motion, the centre of oscillation or percus- 
sion is distant from that end § the length of the line, cylinder or 
prism. 

The axis of motion being in the vertex of the figure, and the figure 
mowing Jlatwise, the centre of oscillation or percussion is distant from 
the vertex, 

In an isosceles triangle, f its height ; 
In a circle, |- its radius ; • 
In a parabola, -f- its height. 

But the axis of motion being in the vertex of the figure, and the 
figure moving sidewise, the centre of oscillation or percussion is dis- 
tant from the vertex, 

In a circle,, f its diameter ; 

3 arc X radius 
In a sector of a circle, j— r — ~i ; 

In a parabola, -f- height + J parameter ; 

radius of base 2 
In a cone, f axis + g-^ ; 

radius 2 X 2 

In a s P here ' radius X 5" + radmS ; . 
In a rectangle, suspended by one angle, § the diagonal ; 
In a parabola, suspended by the middle of its base, f- axis + \ 
parameter ; 

radius 2 X 2 
In a sphere, suspended by a thread, 5 x (radius + i en gth of thread) 
-f- radius + length of thread. 

Example. — Required the length of a rod of uniform diameter 
and density, that, being suspended by one end, will vibrate seconds 
in the latitude of New York. 

2:3:: 39.10153 : 58.6523 inches. Ans 



CENTRE OF OSCILLATION AND PERCUSSION. 297 

Example. — A ball of 4 inches radius is suspended by a string 24 
inches in length ; required the centre of oscillation or percussion in 
the system. 

4 2 X 2 32 

(4-1-24) x~5 = 140 = ' 2286 + 4 +24 = 28.2286 inches from 
the point of suspension. Ans. 

Example. — Required the centre of oscillation in a parabola whose 
height is 10 inches and base 8 inches, supposing the parabola to be 
vibrating side wise. 

10 X 5 4 2 

7 — ~^~ 1 V 3 == ^*676 inches from vertex. Ans. 

The centre of oscillation in a sphere suspended by a point in its circum- 
ference being given, to find the radius of the sphere. 

2h 
h — -=r =s radius ; h being the distance from the point of suspen- 
sion to centre of oscillation. 

Example. — Required the radius of a sphere, that, being suspended 
by a point in its surface, will vibrate seconds at 23° of latitude. 



39.01206 



39.01206 X2 = 9i7S , HiAg Angm 



The radius of the ball or bob of a pendulum being given, to find the 
length of the rod or string by which that ball must be suspended, in 
order that the pendulum may vibrate seconds at a given locality. 

A 2 X 2 \ 
P — I p y r J -j- r= I; V being the length of a pendulum that 

vibrates seconds in the respective locality, r the radius of the ball, 
and / the length of the string or rod ; the last supposed to be with- 
out weight. 

Example. — The radius of a ball being 4 inches, required the 
length of a thread, (supposed without weight,) whereby to form a 
pendulum with that ball, that will vibrate seconds in the latitude 
of New York. 

39.10153 — ( — 4 X 2 f- 4^ == 34.937 inches. Ans. 

\39. 10153 X 5 ' J 



298 CENTRE OF GYRATION. 

CENTRE OF GYRATION. 

The centre of gyration is that point in a revolving body or system 
of bodies, in which, if the whole quantity of matter were collected, 
the angular velocity would be the same ; that is, the momentum or 
quantity of motion in the revolving mass is centred at this point. 

The centre of gyration is at a greater distance from the axis of 
motion than the centre of gravity, and at a less distance from that 
axis than the centre of percussion. Its distance from that axis is a 
mean proportional between the two, in the same mass. 

To find the centre of gyration in a body or system of bodies. 

Rule 1= — Multiply the distance of the centre of percussion from 
the axis of motion, by the distance of the centre of gravity from the 
axis of motion, and the square root of the product will be the dis- 
tance of the centre of gyration from the axis of motion. 

Rule 2. — Multiply the weight of the several particles by the 
squares of their distances from the axis of motion, and divide the 
sum of the products by the sum of the weights ; the square root of 
the quotient will be the distance of the centre of gyration from the 
axis of motion. 

Example. — On a lever (supposed without weight) revolving 
about one end, there are placed — one weight of 4 lbs., at 3 feet 
from the axis of motion ; one of 6 lbs., at 4 feet ; and one of 5 lbs., 
at 5 feet from that axis ; required the centre of gyration in the 
system. 

,/4X3^+ (6_Xi!) + (5 X 52) \ . 

V I 2 l _i_ a _l 5 ) =4.14 feet from the axis 

of motion. Ans. 

A weight of 15 lbs., therefore, placed 4.14 feet from the axis of 
motion, and revolving in the same time, would have the same im- 
petus or momentum as the three weights in their respective places. 

Example. — A lever 6 feet in length, and weighing 14 lbs., is 
revolving about one end ; at 4 feet from the axis of motion, on the 
lever, is placed a weight of 5 lbs., and at 6 feet from that axis, 
another of 8 lbs. ; required the centre of gyration in the system. 

V ( l4Xt'+(SX4-) +j8xj^_ iA55 feet from _ th0 
\ 14 -4- 5 -4- 8 / 

axis of motion. Ans. 

Example. — Required the centre of gravity and centre of percus- 
sion in the last mentioned system. 



CENTRE OF GYRATION. 299 

14X«'+(5X4P) + (8xy) centre g> ^, 

14XI+ (5X4) + (8X6) 

Example. — The centre of gravity in a body, or system of bodies, 
being 4 feet from the axis of motion, and t*he centre of percussion 6 
feet ; at what distance from that axis is the centre of gyration ? 

V (4 X 6) = 4.899 feet. Ans. 

The centre of gyration is distant from the axis of motion : — 
In a straight, uniform rod, revolving about one end ; the length 



//length of rod 2 \ 



of the rod X .57735, or V I — s-g J. 

In a circular plate, revolving on its centre, or a cylinder, revolv- 
ing about its axis, or a wheel of uniform thickness, revolving about 
its axis ; the radius X .7071. 

In a circular plate, revolving about one of its diameters as an 
axis ; the radius X -5. 

In a thin, hollow sphere, revolving about one of its diameters as 
an axis ; the radius X .8164. 

In a cone, revolving about its axis ; the radius of the base X 
.5477. 

In a four-sided equilateral pyramid, revolving about its apex ; 
the axis X .866. 

In a paraboloid, revolving about its axis ; the radius of the base 
X -57735. 

In a sphere, revolving about one of its diameters as an axis ; the 
radius X -6325. 

In a straight lever, the arms being R and r, = a//p~ , o 

In wheels in general revolving about their axes, = 



,/R X r 2 X 2 + A X I' 2 X 2\ 
W I /p I a \ y o ) ; R being the weight of the 

m, r the radius of the wheel, I the length of the arms, and A the 
eight of the arms. 
In a water-wheel in operation, = 
,/RXr2X2 + AX/ 2 X2 + WXrA , _ 

v V (R + a + w) x 2 > nearl y ' W bein S 



the weight of the water on loaded arch. 



300 CENTRAL FORCES. 

The Energy of a revolving body = 

Velocitv of centre of gyration in feet per second 2 

%£ " X weight of 

revolving body. 

The energy of a revolving body is its weight multiplied into the 
height due to the velocity with which the centre of gyration moves 
in its circle. The height in feet, due to a velocity in feet per second 
«= (velocity -J- 8.02) 2 , or velocity 2 -J- 64£. 

Example. — A grindstone two feet in diameter and weighing 140 
lbs., makes 80 revolutions per minute ; required the energy with 
which the stone moves, or the mechanical power that must be com- 
municated, to give it that motion. 

(2 X -7071 X 3.1416 X 80\ 2 

I gn ; ) -r- 64J = .5456, energy, the weight 

being 1, and .5456 X 140 = 76.38 lbs. Ans. 



CENTRAL FORCES. 

The Central Forces are the centrifugal force and centripetal force ; 
these forces are opposed to each other. The former is the force with 
which a body, moving in a curve, tends to fly off from the axis of its 
motion, and the latter is the force that maintains the body in its 
curvilinear path. 

To find the centrifugal force of a body. 

Revolutions per minute 2 X diameter of centre of gyration's circle in feet 

5865" = Centnfil - 

gal force, the weight being 1. 

/Velocity of centre of gyration in feet per second\ 2 . 

I ■— ; ) -r- twice the distance of the centre 

v 4.01 ' 

of gyration from centre of motion = centrifugal force, the weight being 1. 

If we let 
v represent velocity of centre of gyration in feet per second, 
r " radius, in feet, of circle described by centre of gyration, 
w " weight of body, 
c " centrifugal force ; then — 

v 2 X w 



rX-32* 

v 2 X™ 
cX32£ 



= c. 



cX32£Xr_ 



w. 



v (iXMX5)= : 



FLY-WHEELS. THE GOVERNOR. 



301 



Example. — Required the centrifugal force of a fly- wheel whose 
centre of gyration is 8 feet from the axis of motion, and which centre 
moves with a velocity of 34 feet per second. 

( — ) -i- (8 X 2) = 4.49 times the weight of the wheel. Ans. 



In fly-wheels in general, 

Diameter of wheel in feet X -6136 
time in seconds of one revolution 



X weight of rim = centrifugal force, nearly. 



FLY-WHEELS. 

Fly-wheels are used both as regulators of force and magazines of 
power; when, more especially for the former purpose, they should 
be placed as near as practicable to the prime mover; and when, 
more especially for the latter, as near as practicable to the working 
point. 

To find the weight of the rim of a fly-wheel proper for an engine of a 

given horse-power. 

Horse-power of engine X 1368 . . ... , . . - ._ 

: = requisite weight of run in 100 

diameter of wheel in feet X revolutions per minute 

pounds. 

Example. — What should be the weight of the rim of the fly-wheel 
of a 40 horse-power engine, the wheel being 16 feet in diameter, and 
making 42 revolutions per minute ? 

40 X 1368 
16x42 == 81.43 X 100 = 8143 lbs. Ans. 



THE GOVERNOR. 

The Governor is a regulator of force merely, and acts upon the 
principle of central forces. It makes half as many revolutions in 
any given time, as a pendulum, the length of which is equal to the 
perpendicular distance between the point at which the arms are 
suspended and the plane in which the centre of oscillation in the 
system moves, makes in the same time. 

If we let 

I = perpendicular distance referred to, in inches, 

r sss number of revolutions made by the system per minute, 

Then, 

/ 187.6 V = j -i 187.6 



and 



26 V r J ' */l 



302 FORCE OF GRAVITY. 

Example. — A governor makes 40 revolutions a minute ; what is 
the perpendicular distance from the point at which the arms are 
suspended, to the plane in which the centre of oscillation moves ? 

187.6 -^ 40 = 4.69 X 4.69 = 22 inches. Ans. 
\/lX -31986 = time in seconds of each revolution. 



FORCE OF GRAVITY. 

The Force of Gravity is greatest at the earth's surface, and de- 
creases from the surface to the centre as the distance from the sur- 
face increases ; from the surface upwards it decreases as the square 
of the distance from the centre, in semi-diameters of the earth. 

The semi-diameter is usually taken at 4000 miles. 

A body, therefore, weighing at the earth's surface 500 lbs., 
would, were it removed 1000 miles from the surface towards the 
the centre, weigh 

4000 : 500 : : 3000 : 375 lbs. 

And were the same body placed 1000 miles above the earth's sur- 
face, it would be 

(4000 + 1000) -7- 4000 = l£ semi-diameters from the centre, and 
at that locality would weigh — 

500 -f- 1.252 = 320 lbs. 

And in order that the same body may weigh but 250 lbs., it must 
be elevated above the surface 

/v/(500 ~r- 250) = 1.4142 semi-diameters from the centre, or 
(4000 X 1.4142)— 4000 = 1657 miles. 

Note. — The force of gravity at any locality, as measured by the descent of a body 
falling freely through space, is equal in feet per second to the length of a pendulum in 
feet that vibrates seconds at that locality X 4.93483, for the first second of the body's de- 
scent. 

Thus, the force of gravity of a body for the first second of its de- 
scent, at the level of the sea, at the "latitude of 41°, is 3.25846 X 
4.93483 = 16.08 feet. 

To find the time which a body will be in falling from rest, through any 
space, the space fallen through being given, or the maximum velocity 
attained in falling being given. 

. /Space fallen through in feet\ 
ji y j — time of falling w seconds. 

16-Ar 



FORCE OF GRAVITY. 303 

Maximum velocity attained in feet per second 



^ 



= time of falling in seconds. 



Space fallen through m feet X 2 

- = tune of falling m seconds. 

maximum velocity attained m feet per second 

To find the maximum velocity attained by a body falling freely from rest, 
through space, the space fallen through being given, or the time of de- 
scending being given. 

\/ (Space fallen through in feet X 16 12") X 2 = maximum ve- 
locity attained in feet per second. 

Time of falling in seconds X 32^ = maximum velocity attained in 
feet per second. 

Space fallen through in feet X 2 



time of falling in seconds 



= maximum velocity attained in feet per second. 



To find the space fallen through by a body falling freely from rest, the 
time of descending being given, or the maximum velocity attained in 
falling being given. 

Time of falling in seconds 2 X 16 tV — space fallen through in feet. 

Maximum velocity attained in feet per second 2 ' „ 1T . , . , L 
- = space fallen through m feet. 

Maximum velocity attained in feet per second X h time of falling 
in seconds = space fallen through in feet. 

To find the height of a stream projected vertically from a pipe. 

Quantity in cub. ft. discharged per minute X 2.4 . . .. . „ 

: — — -— : — - ;— = maximum velocity m feet per 

area of discharging orifice m mches 

second, or velocity in feet per second when the stream leaves the pipe. 
''Maximum velocity of stream in feet per second^ 2 

locity, or height in feet to which the stream will ascend. 

Example. — From a fire-engine there was discharged through a 
pipe | of an inch in diameter, 15 cubic feet of water in 1 minute, the 
pipe being directed vertically ; what height did the stream attain? 

( 15 X 1728 _Y= 



(Maximum velocity of stream in feet per second\ 2 , . , . 
— — : J = height in feet due to that ve- 



.75 X .75 X .7854 X 12 X 60 X 8. 



103.22 feet. Ans. 



To find the force or power requisite to produce the foregoing result. 
Height of column of water equal to pressure of atmosphere (33.87 feet) 



pressure of atmosphere on 1 square inch of surface (14.7 lbs.) 
eolumn of water of 1 in:h transverse area, weighing 1 lb. = 2.305 feet 5 then. 



= height of 



304 PENDULUMS. 

Pounds of water discharged per second X 2.305 

77 — z = "^ — : — : — ; = maximum velocity in feet pei sec- 
area of dischargmg orifice in niches 

ond, and 

Maximum velocity in feet per second X weight in lbs. discharged 
per second = momentum in lbs., or force in lbs. required. 

V (Height of projection in feet X 16*) X 2 = maximum veloc- 
ity in feet per second. 

Example. — Required the constant force requisite to project a con- 
tinuous stream of water perpendicularly to the height of 103.22 
feet, through a pipe | of an inch in diameter. 

V (103.22 X16 T V)X 2 
2.305 -f Tfb * x ". 7854 X V (103.22 X 16*) X 2 = 1275 lbs. Arts. 

15 X 62.5 X 2.305 15 X 62.5 
' 60 X -75 2 X .7854 X 60 "~ 1275 lbs " 

Note. — The force requisite to project a body vertically to any given height is equal to 
the force or momentum with which that body will return to the place from whence pro- 
jected. The time occupied in ascending is equal to that occupied in descending. 



OF PENDULUMS. 

Pendulums of the same length vibrate faster the farther they are 
removed from the equator, either north or south. 

The length of a pendulum to vibrate seconds, or 60 times in a 
minute, at the level of the sea, at the temperature of 60° F., at the 
latitude of Trinidad, is 39.01879 inches; at Jamaica, 39.03508 ; at 
New York, 39.10153 ; at Bordeaux, 39.11282; at Paris, 39.12843 ; 
at London, 39.1393; at Edinburgh, 39.1554; and at Greenland, 
39.20328 inches. 

To find the length of a pendulum that will make a given number of 
vibrations in a given time. 

Rule. — As the number of vibrations required in one minute is to 
60 vibrations, so is the square root of the length of a pendulum that 
makes 60 vibrations in a minute at the respective locality, to the 
square root of the length of the pendulum required. 

Example. — Required the length of a pendulum that will vibrate 
half seconds at the level of the sea, at the latitude of New York. 

120 : 60 : : V39.10153 : 3.126561, and 
3.126561 X 3.126561 = 9.77538 inches. Ans. 



SCREW-CUTTING IN A LATHE. 305 

To find the number of vibrations that a pendulum of a given length will 
make in a given time. 

Rule. — As the square root of the length of the given pendulum 
is to the square root of the length of a pendulum that vibrates sec- 
onds in the respective locality, so is 60 vibrations to the number of 
vibrations it will make per minute. 

Example. , — Required the number of vibrations that a pendulum 
24 inches in length will make per minute, at the level of the sea, at 
the latitude of New York. 

V24 : V39.10153 : : 60 : 76.585. Ans. 

To find the length of a pendulum that will vibrate as many times in a 
minute as there are inches in that pendulum 1 s length. 

Rule. — Multiply the square root of the length of a pendulum 
that vibrates seconds in the respective locality, by 60, find the cube 
root of the product, and the square of the last root will be the length 
sought. 

Example. — What must be the length of a pendulum, in order that 
its length in inches and the number of its vibrations per minute, at 
the latitude of New York, may be equal ? 

V (39.10153) X 60 = 375.1873, and 
4/375.1873 = 7.2125, and 
7.2125 2 == 52.02 inches. Ans. 

Note. — The length of a pendulum is the distance from the point of suspension to the 
centre of oscillation. See Centres of Oscillation and Percussion. 



SCREW-CUTTING IN A LATHE. 

Screws to any degree of fineness not exceeding about \ inch pitch, 
that is, having not more than about 8 threads in an inch, may be cut in 
a lathe by means of a wheel on the end of the leading screw, and an- 
other on the end of the lathe spindle, with a carrier, or intermediate 
wheel, called a stud-wheel, to connect them. And all that is neces- 
sary in this kind of training, in order to obtain a screw of any given 
or predetermined pitch, is, that the wheel on the end of the leading 
screw bear the same ratio to that on the end of the lathe spindle, 
that the pitch of the screw employed bears to that of the screw 
intended. 

The carrier, whatever be the number of its teeth, and the same 
would be true if two or more carriers, of whatever size, were employed, 
26* 



306 SCREW-CUTTING IN A LATHE. 

in no way affects the ratio of velocity between the wheels alluded 
to. It receives the motion directly from one of them, and conveys 
it directly to the other, and is introduced to obviate the necessity of 
employing larger wheels. It is both a driven and driver in the train, 
and, consequently, in calculating the gearing to produce a screw of 
a given or required pitch, needs not to be taken into account. 

But screws of a greater degree of fineness than about 8 threads in 
an inch, more especially if the pitch of the screw attached to the 
lathe be large, are more conveniently cut by help of two intermedi- 
ate wheels, of unequal sizes, both upon the same shaft, called a stud- 
wheel and pinion; whereby the requisite velocity of the spindle may 
be more effectively obtained, and with a nearer approach to uni- 
formity in size of the wheels employed. When this last mode of 
training is resorted to, all the wheels in the train are influential in 
determining the pitch the contemplated screw will take ; and the 
equation will be as the product of the drivers to the product of the 
driven ; that is — 

If we let a = pitch of leading screw, ) 

a' = No. of teeth in stud pinion, \ drivers, 

a" = No. of teeth in spindle- wheel, ) 

b = pitch of contemplated screw, ) 

b' = No. of teeth in stud- wheel, > driven, 

b" = No. of teeth in leading screw-wheel, ) 

We shall have, a X af X «" = b X V X V. 

And, in the train first supposed, whether with or without car- 
riers, — 

aXa" = bXh"; 

And the product of either full side of the equation, divided by the 
product of either two terms in the opposite, will give the remaining 
term, or term sought. 

Example. — a is | inch, a" has 24 teeth, and b is to equal | inch ; 
required the requisite number of teeth in b". 

§ X 24 = 9, and 9 -r- 1 = 36 teeth. Ans. 
Or, (.375 X 24) -^ .25 = 36 ; or, .25 : 24 : : .375 : 36. Ans. 

Example. — a" has 60 teeth, a' has 20, and a = £ inch ; b" has 120 
teeth, and b is fixed upon at y 1 ^ inch ; required the number of teeth 
requisite for V. 

(£ X 20 X 60) -j- (120 X tV) = 80 teeth in V. Ans. , 



Or, .5 X 20 X 60 -j- 120 X -0625 = 80. Ans. 



SCREW-CUTTING IN A LATHE. 307 

But it is more convenient, frequently, in solving problems of this 
nature, to substitute the number of threads in a given length of the 
screws, for their pitches, respectively. 

Thus, if we let a = No. of threads in an inch of leading screw, 

b = No. of threads in an in. of contemplated screw, 

We shall have, as an equation for the foregoing train — 

aXV XV = b X«' X a". 

That is, the product of the teeth in the driving-wheels multiplied 
by the number of threads in an inch of the contemplated screw, is 
equal to the product of the teeth in the driven wheels multiplied by 
the number of threads in an inch of the leading screw. 

Example. — a' has 20 teeth, a" has 60, and b is to have 16 threads 
in an inch ; a has 2 threads in an inch, and b" has 120 teeth ; re- 
quired the number of teeth requisite for V . 

20 X 60 X 16 

120 X 2 = 80 ' Ans - 

Example. — The leading screw has 2 threads in an inch, the lead- 
ing screw-wheel has 120 teeth, the stud- wheel 80 teeth, and the 
spindle-wheel 60 teeth ; required the number of teeth that must be 
in the stud pinion, in order that the contemplated screw may have 
16 threads in an inch. 

% 80 

% X *$0 X 80 = 4~ = 20 teeth in a'. Ans. 

~WxW~ 



The following table exhibits the change wheels proper for cutting 
screws of various pitches from 1 inch to -^ inch, or from one thread 
in an inch to 36 threads in an inch, the leading screw having a pitch 
of £ inch. 

Note. — If the leading screw have 3 threads in an inch, the following table is made ap- 
plicable by changing either of the driven wheels for one having I less number of teeth. 
3 •. 2 : : tabular driven : driven required, when the leading screw has § inch pitch, and 4 : 
2 : : tabular driven : driven required, when the leading screw has i inch pitch, &c. Or 
2:3:: either of the tabular drivers : driver required in its stead, when the leading screw 
has 3 threads in an inch, and 2:4:: either of the tabular drivers : driver required in its 
stead, when the leading screw has 4 threads in an inch, &c. 

Pitch of leading screw X No. teeth in stud-wheel X No. teeth in screw-wheel = pitch of 
contemplated screw X No. teeth in stud-pinion X No. teeth in spindle-wheel, in all 
instances. 

The tabular wheels may be taken in any proportion, greater or less. 



308 



STEAM AND THE STEAM ENGINE. 



TABLE 



Of Change Wheels for Screw-cutting ; the leading screw being of £ 
inch pitch, or containing 2 threads in an inch. 



■s 

•9 . 


Number of 
teeth in 


•9 I 


Number of teeth in 


1. 

a is 


Number of teeth in 






















■3 


H 




■<3 






4 


•« a> 


"S 






01 


S3 c3 

■So 


.3 

te 


i 


1« 


1 






is 


5a 


1 






i 


I 

3 


is 

1 

bo 

1 
c3 

CD 
H3 


II 

|! 

§8 
fc"S 


0) 
n3 
.9 
ft 
tn 

! 


is 

1 

Xfl 


1 
I 

02 


is 

u 

8 

If 

'i 

hi 


Si 

*! 

s ° 

g o 


0) 

1 
ft 

1 


is 

t/2 


1 
'ft 

1 


u 

8 

o° 
•■B 

S3 
03 

H3 


1 


80 


40 


84 


40 


55 


20 


60 


19 


50 


95 


20 


100 


U 


80 


50 


si 


90 


85 


20 


90 


194 


80 


120 


20 


130 


14 


80 


60 


8<| 


60 


70 


20 


75 


20 


60 


100 


20 


120 


it 


80 


70 


84 


90 


90 


20 


95 


204 


40 


90 


20 


90 


2 


80 


80 


9| 


40 


60 


20 


65 


21 


80 


120 


20 


140 


2| 


80 


90 


10 


60 


75 


20 


80 


22 


60 


110 


20 


120 


24 


80 


100 


104 


50 


70 


20 


75 


224 


80 


120 


20 


150 


2| 


80 


110 


11 


60 


55 


20 


120 


22| 


80 


130 


20 


140 


3 


80 


120 


12 


90 


90 


20 


120 


23! 


40 


95 


20 


100 


3| 


80 


130 


12| 


60 


85 


20 


90 


24 


65 


120 


20 


130 


H 


80 


140 


13 


90 


90 


20 


130 


25 


60 


100 


20 


150 


3| 


80 


150 


134 


60 


90 


20 


90 


254 


30 


85 


20 


90 


4 


40 


80 


131 


80 


100 


20 


110 


26 


70 


130 


20 


140 


4| 


40 


85 


14 


90 


90 


20 


140 


27 


40 


90 


20 


120 


4i 


40 


90 


1*4 


60 


90 


20 


95 


274 


40 


100 


20 


110 


4| 


40 


95 


15 


90 


90 


20 


150 


28 


75 


140 


20 


150 


5 


40 


100 


16 


60 


80 


20 


120 


284 


30 


90 


20 


95 


54 


40 


110 


164 


80 


100 


20 


130 


30 


70 


140 


20 


150 


6 


40 


120 


16J 


80 


110 


20 


120 


32 


30 


80 


20 


120 


64 


40 


130 


17 


45 


85 


20 


90 


33 


40 


110 


20 


120 


7 


40 


140 


174 


80 


100 


20 


140 


34 


30 


85 


20 


120 


74 


40 


150 


18 


40 


60 


20 


120 


35 


60 


140 


20 


150 


8 


30 


120 


18| 


80 


100 


20 


150 


36 


30 


90 


20 


120 



OF STEAM AND THE STEAM ENGINE. 



It has been found by experiment that the force requisite to over- 
come the friction of a locomotive engine and attendant machinery, 
the engine being without load, is equal to a pressure of about 1.15 
lbs. effective on each square inch of the cylinder's cross sectional 



STEAM AND THE STEAM ENGINE. 309 

area, or equal to 1 lb. effective for the engine, and 0.15 lb. effective 
for the attendant machinery ; this item, however, for practical pur- 
poses, is usually taken at 1£ lbs. ; by effective pressure is meant a 
pressure over and above that of the atmosphere, or over and above 
14.7 lbs. on each square inch of surface. 

If we let 

d = diameter of cylinder in inches, 

s = stroke of piston in feet, 

r = revolutions per minute, 

p = mean effective pressure in lbs. per square inch, as shown by 
the indicator, 

a = sXrX2 = velocity of piston in feet per minute, 

h = 1^ lbs. = effective pressure on each square inch of cylinder's 
cross sectional area, requisite to overcome friction, then, 

To find the effective power or force of a steam engine. 

d 2 X (p — h) X a 

42017 = e ^ ec ^ ve fo rce * n horse-power. 

Example. — "What is the effective force of a steam engine, the 
diameter of the cylinder (d) being 36 inches, the stroke of the piston 
(5) 7 feet, the effective pressure (p) 30 lbs., and making 17£ revolu- 
tions (r) per minute ? 

36 2 X (30 — 1.5) X 7 X 17.5 X2r 42017 = 215.37 horse- 
power. Ans. 

To find the nominal power of a low-pressure or condensing engine. 

d*Xa 



6000 



nominal force in horse-power. 



To find the nominal power of a high-pressure, or non-condensing 
engine. 
d 2 XpX a 
120 000 = nom ^ na ^ f° rce * n horse-power. Or, 

* xpxV s . ; . . _ ;. . + 

cyT) == nominal torce m horse-power, the piston moving 

at the ordinary speed of 128 times the cube root of the stroke. 

To find the pressure of the steam on each square inch of the boiler's 
surface. 

Let P = pressure in lbs. on safety valve, 

p = pressure in lbs. on each square inch of surface, 
W = weight or resistance in lbs., as indicated by the spring 
balance, * 






310 STEAM AND THE STEAM ENGINE. 

w = sum of the weights of the lever, safety valve, and bal- 
ancing weight, in lbs., 

s == length of lever in inches from its axis of motion to a 
point vertical to the centre of the valve, 

I = length of lever in inches from its axis of motion to W, or 
to the point on the lever at which the spring balance is attached, 
a = area of safety valve in square inches. 
Then, 

P — w -{- w 

s : / : : W : P — w ; and = p. 

a r 

Example. — The length of the lever from its axis of motion to a 
point over the centre of the safety valve (s) is 3 inches, its length 
from the axis of motion to the point at which the weight or spring 
balance is attached (I) 24 inches, the weight, or pressure, as indi- 
cated by the spring balance (W) 40 lbs., the sum of the weights of 
the lever, valve, &c, (w) 6 lbs., and the area of the valve 6£ square 
inches ; required the pressure of the steam per square inch. 

3 : 24 : : 40 : 320, and 
320 + 6 = 326 -r- 6.5 = 50.15 lbs. Ans. 
W : P — w : : s : I Z : s : : P — w :W. 
V — w:W::l:s. 

To find the volume of steam compared with the volume of water. 

Let F = elastic force of steam in pounds per square inch, 
V = volume of steam compared with volume of water. 
Then, 

24250 . 24250 

— =p— -j- 65 = V, and y nr = F, nearly ; the volume of water 

in every instance, being 1. 

Example. — The elastic force of the steam is 40 lbs. to each square 
inch of the surface of its bulk ; what space does it occupy, compared 
with the space it would occupy if it were condensed to water ? 

24250 -j- 40 = 606 -f 65 = 671 ; that is, when the elastic force 
of the steam is 40 pounds to each square inch of the surface of its 
bulk, the volume of the steam, compared with its volume in water, 
is as 671 to 1. 

Note. — The preceding formulas are not strictly correct for all densities of steam, but 
give the mean of a general range. 

The volume of steam compared with the volume of water as 1, has been found, by ex- 
periment, when the elastic force is 14.1 lbs. = 1694 ; 20 lbs. = 1280 ; 25 lbs. = 1044 ; 
30 lbs. = 883 5 35 lbs. =s Y67 ; 40 lbs. = 679 ; 45 lbs. = 610 ; 50 lbs. = 554 ; 55 lbs. = 
508 ; 60 lbs. = 470 ; 65 lbs. = 437 ; 70 lbs. = 408 ; 75 lbs. = 383 ; 80 lbs. = 362 •, 90 
lbs. == 325 ; 100 lbs. = 295 ; 150 lbs. = 205 ; 200 lbs. — 158. 



STEAM AND THE STEAM ENGINE. 311 

To find the temperature of the steam, its elastic force being known. 

Rule. — Multiply the 6th root of the elastic force in inches of 
mercury by 177, and subtract 100 from the product ; the remainder 
will be the temperature in degrees, Fahrenheit nearly. 

Or, if we let 

a = elastic force in inches of mercury, 

b = temperature of steam in degrees, F. 
Then, 

B, /& + 100\ 6 

A/ a X 177 - 100 = b, and ^ -^r- J = a. 

Example. — The elastic force of the steam is 150 inches of mercury, 
(150 -J- 2.04 = 73£ pounds to each square inch of the surface of its 
bulk). ; what is the temperature ? 

/^150 = 2.305 X 177 = 408 — 100 = 308°. Ans. 

Note. — The pressure of the steam in pounds per square inch X 2.04 = pressure in 
inches of mercury. 

The temperature due to a pressure of 14.7 lbs. per square inch = 212° ; 20 lbs. = 228° ; 
25 lbs. = 241° 5 30 lbs. = 252° ; 35 lbs. = 261° ; 40 lbs. = 269° ; 45 lbs. = 276° ; 
50 lbs. = 283° ; 55 lbs. = 289° ; 60 lbs. = 296° ; 65 lbs. = 301° ; 70 lbs. = 306° ; 75 
lbs. = 311° : 80 lbs. = 316° ; 85 lbs. = 320° ; 90 lbs. = 324° ; 100 lbs. = 332° ; 150 
lbs. = 363° j 200 lbs. = 387°. 

To find the quantity of water required for steam per minute by an engine 
in motion. 

Rule. — Multiply the velocity of the piston in feet per minute, by 
the square of the cylinder's diameter in feet multiplied by 0.7854, and 
divide the product by the volume of steam compared with the vol- 
ume of water, due to the pressure exerted. 

Example. — The diameter of the cylinder is 2^ feet, the velocity 
of the piston 245 feet per minute, and the constant pressure exerted 
by the steam 60 pounds to the square inch ; what quantity of water 
must be converted into steam per minute ? 

2.5 2 X .7854 X 245 -7- 470 = 2.56 cubic feet. Ans. 

Note. — For a single-acting engine half the quantity indicated by the above rule is re- 
quired. 

To find the quantity of steam required to raise a given quantity of 
water of a given temperature to a required temperature. 

For the present purpose, we may assume that the sum of the la- 
tent and sensible heat of vapor of water is constant, and= 1178°. 
Therefore, the latent heat of vapor of water at 32° = 1 1 78 — 32 



312 STEAM AND THE STEAM ENGINE. 

= 1146; at 100° = 1178 — 100=1078°; at 212° = 1178 — 212 
= 966°, &c. 

The latent heat of water at 32° of sensible heat = 140°. 

If we let 

a = temperature of steam employed, 

b = temperature of water to be raised, 

c = temperature to which the water is to be raised, 

d = volume of steam compared with the volume in water due to 
the temperature of the steam, 

w = quantity of water in cubic feet to be heated ; 

Then 
c 

oj,, — — — = quantity of water (say in cubic feet) that must 

be converted into steam having a temperature a, required to raise 
1 cubic foot of water from b to c. 

(c — b) X d 

^ J = quantity of steam in cubic feet, at temperature a, 

ybb — t— (X — c 

required to raise 1 cubic foot of water from b to c. 

w \s (c b) 

p - = quantity of water in cubic feet that must be con- 

966-|-a — c H J 

verted into steam, and have a temperature at a, required to raise 
the given quantity of water from b to c. 

w ^ ' ' = quantity of steam in cubic feet, at tempera- 

966-}-a — c ^ J i if 

ture a, required to raise the given number of cubic feet of water 

from b to c. 

Example. — What quantity of water in steam at 212° will raise 
100 cubic feet of water from 60° to 200° ? 

10 ° X U ° = 14.31 cubic feet. Ans. Making 14.31 X 

966 + 212 — 200 ^ ~ 

1694 = 24241 cubic feet of steam at 212 degrees. 

To find the quantity of water at a given temperature required to 
reduce a given quantity of steam to a given temperature. 

Let 

a — temperature of steam to be reduced, 

c = temperature to which the steam is to be reduced, 

b == temperature of the water injected. 

Then 

' , =. number of cubic inches of the water required to 

c — b 

reduce 1 cubic foot of the steam from a to c nearly. 



STEAM ACTING EXPANSIVELY. 



313 



To find the Mean Force of Steam acting Expansively , ana the Advan- 
tage Gained. 



N. 


C. 


N. 


C. 


N. 


C. 


N. 


C. 


N. 


C. 


1.1 


1.095 


2.1 


1.742 


3.1 


2.131 


4.2 


2.435 


6.3 


2.841 


1.2 


1.182 


2.2 


1.788 


3.2 


2.163 


4.4 


2.482 


6.5 


2.872 


1.3 


1.262 


2.3 


1.833 


3.3 


2.194 


4.6 


2.526 


7. 


2.946 


1.4 


1.336 


2.4 


1.875 


3.4 


2.224 


4.8 


2.569 


7.5 


3.015 


1.5 


1.405 


2.5 


1.916 


3.5 


2.253 


5. 


2.609 


8. 


3.079 


1.6 


1.470 


2.6 


1.956 


3.6 


2.281 


5.2 


2.649 


8.5 


3.140 


17 


1.541 


2.7 


1.993 


3.7 


2.308 


5.4 


2.686 


9. 


3.197 


1.8 


1.588 


2.8 


2.030 


3.8 


2.335 


5.6 


2.723 


10. 


3.303 


1.9 


1.642 


2.9 


2.065 


3.9 


2.361 


5.8 


2.758 


11. 


3.398 


2. 


1.693 


3. 


2.099 


4. 


2.386 


6. 


2.792 


12. 


3.485 



S = stroke of piston in inches. 

d = distance the piston moves in inches before the steam is cut off. 
P = pressure in pounds per square inch of the steam on the piston. 
C = tabular quantity in column C, standing against the quantity in column N, that 
= S -r d. 

When S -r- d = any quantity in column N (above table), then, if 
S -7- d be taken to represent the whole effect of the steam, or effect 
that would have been produced had the steam not been cut off, the 
quantity in column O (same table) , standing against the quantity 
S -f- d, will represent the effect, the steam being cut off as supposed ; 
that is, the effect of the steam, cut off as supposed, will be to what 
it would have been had no cut-off taken place, as the quantity in 
column C to the quantity in column N, against it. 

CJ?d 

-h— = CXPt(S-t^) = mean pressure of steam in pounds 

per square inch on the cylinder, the steam being cut off ad libitum. 

Example. — The steam enters the cylinder with a force or pressure 
of 40 pounds to the square inch, is cutoff when the piston has moved 
32 inches, and the whole stroke of the piston is 74 feet (87 inches ) ; 
required the mean pressure per square inch on the cylinder. 

87 -r- 32 = 2.7, the pressure, therefore, is to what it would have 
been, if the steam had not been cut off at all, as 1.993 to 2.7, as yVVu i 
and the mean force or pressure on the cylinder is 

1.993 X 40 X 32 — 87 = 29.3 lbs. per square inch. Ans. 



Steam, under a pressure of 1 atmosphere, flows into a vacuum with a velocity of about 
1400 feet a second, and into the air with a velocity of about 650 feet a second ; its velocity 
under a pressure of 20 atmospheres is about 1600 feet a second, either into the air or a 
vacuum. 

Atmospheric air at 60°, b. 30 in., flows into a vacuum with a velocity 1327 feet a second. 

The velocity of a cannon ball (maximum) is about 2000 feet a second, when near the muzzle 
of the gun, but at the distance of 500 yards from the gun it is not above 1300 feet a second. 

The greatest velocity of a rifle ball, near the muzzle of the gun, is 2012 feet a second. 

The velocity of a musket ball, full charge, 18 balls to the pound, windage .05, near the 
muzzle of the gun, is 1600 feet a second. 

Inflamed gunpowder expands with a velocity of about 5000 feet a second. 
27 



314 OF CONTINUOUS CIRCULAR MOTION. 



OF THE ECCENTRIC IN A STEAM ENGINE. 

The throw of the eccentric is equal to twice the distance from the centre of 
formation to centre of revolution ; that is, it is equal to the diameter of the 
circle of revolution minus the diameter of the circle of formation. If r rep- 
resent the shortest, and R the longest, of all the radii in the eccentric, meas- 
uring from the centre of the circle of formation or axis of the revolving shaft; 
then (R — r) X 2 = the throw of the eccentric. 

The travel of the valve is equal to the sum of the widths of the two steam 
openings, plus a slight excess of length to the valve more than just sufficient 
to cover the openings. 

L = length of lever in inches on weigh or traverse shaft for working the 
valve. 

/ = length of lever in inches on weigh-shaft for eccentric rod. 

v = travel of valve in inches. 

t = throw of eccentric in inches. 

vl -i- L = t ; vl-r- t = h; ht -f- l = v ; ~Lt -±-v = l. 

To find the number of revolutions that each of two wheels that are geared 
together will make before the same teeth will come together again. 
Rule. — Divide the numher of teeth in each wheel by the greatest num- 
ber that will divide both without a remainder ; the greater quotient will be 
the number of revolutions made by the smaller wheel, and the less the num- 
ber made by the larger. If both wheels cannot be divided by a common 
divisor, the smaller wheel will make as many revolutions as there are teeth 
in the larger, and the larger as many as there are teeth in the smaller. 

Example. — Required the number of revolutions made by each of 
two wheels that are geared together, the larger having 66 teeth, and 
the smaller 21, before the same teeth come together again. 
66 22 revolutions of smaller wheel. ) 

21 * ~ 7' revolutions of larger wheel. ) 



OF CONTINUOUS CIRCULAR MOTION. 

When a series of wheels, or wheels, pinions, drums and pulleys, 
are so arranged that one, being set in motion, imparts motion 
directly to another, that to a third, and so on, all, at equal distances 
from their respective centres, will describe equal circles of gyration. 
As are their radii, diameters, circumferences or number of teeth, 
therefore, one to another, so are their number of revolutions one to 
another, or so are their turns in the same space of time. 

In every machine there is some first point of impulse, or point at 
which the motive power is applied, and the circle of gyration de- 
scribed by that point is the periphery of the first moved, or first 
driven, of that machine. From some point in this circle, or from 
some circle described upon the shaft which it drives, the power is 
transmitted to a remoter or next contiguous movement, thereby the 
point so transmitting becoming the driver thereof. Thus the whole con- 
tinuous chain consists, alternately, of driven and drivers throughout ; 



CIRCULAR MOTION. 315 

or, if we take the motive power into account of drivers and driven 
throughout. 

Example. — A drum, on the main line of shafting-, is 18 inches in 
diameter, and by means of a belt drives a pulley whose diameter is 12 
inches ; how many revolutions does the pulley make, to one revolution 
of the drum ? . 

18 -j- 12 = 1.5. Ans. 
What portion of a revolution does the drum make, to one revolu- 
tion of the pulley 1 

12 -j- 18 = | of one revolution. Ans. 

Example. — The above drum makes 120 revolutions a minute ; 
how many revolutions does the pulley which it drives make in the 
same time ? 

18 X 120 = 2160 -$- 12 = 180 revolutions. Ans. 
Or, 2 : 3 :: 120 : 180 revolutions. Ans. 

Example. — The diameter of the driver is 18 inches, and makes 
120 revolutions a minute ; what must be the diameter of a pulley which 
it will drive at the rate of 180 revolutions a minute ? 
120 X 18 -7- 180 = 12 inches. Ans. 

Example. — A pinion of 8 teeth drives a wheel of 49 ; how many 
revolutions does the pinion make to one revolution of the wheel ? 
49 -f- 8 = 6J revolutions. Ans. 

Example. — A pinion has 8 teeth, and makes 80 revolutions a 
minute ; how many revolutions does a wheel make, in the same time, 
which has 49 teeth, and works in contact? 

80 X 8 -j- 49 = 13^ revolutions. Ans. 

Example. — A wheel has 49 teeth, and makes 13^. revolutions in 
a given time ; how many teeth must a wheel or pinion have to work 
in contact, and make 80 revolutions in the same time ? 
49 X 13 -^ -r- 80 = 8 teeth. Ans. 

To find the number of revolutions made by the last, to one revolution of 
the first, in a train of wheels and pinions. 

The last wheel, or pinion, in a train, whichever it be, is neeessarily 
a driven ; — therefore, 

Rule. — Divide the product of all the teeth in the drivers by the 
product of all the teeth in the driven ; the quotient is the number or 
ratio sought. 

Example. — A wheel of 72 teeth drives a pinion of 14, upon 
whose shaft is a wheel of 56 teeth that drives a pinion of 10, upon 



316 CIRCULAR MOTION. 

•whose shaft is a wheel of 35 teeth that drives a pinion of 6 ; how 
many revolutions does the last pinion make, to one revolution of the 
first wheel ? 

72 X 56 X 35 

14 V 10 V 6 ™" 168 revolutions. Ans. 

Example. — A wheel of 72 teeth drives another of 29, upon whose 
shaft is a pulley of 24 inches diameter that gives motion to one of 10 ; 
what number of revolutions are made by the last pulley, to one revo- 
lution of the first wheel ? 

72 X 24 

2 Q v 10 == 5.96 revolutions. Ans. 

The rule is Well established, that, in training for the purpose of 
accumulating velocity, or for the purpose of diminishing an over ac- 
cumulated, a certain mean or proportional velocity, between the sev- 
eral movers, should exist ; and, further, that without some important 
reason for a higher ratio, the number of teeth on the wheel should not 
exceed 6 to 1 on the pinion with which it works. 

The mean, or proportional velocity alluded to, is found by the fol- 
lowing rule, and is applied as shown in Examples. 

Rule. — Multiply the given and required velocities together, and 
extract the square root of the product, which is the mean sought. 

Example. — From a wheel of 72 teeth, making 20 revolutions a 
minute, motion is to be conveyed, by help of two intermediate wheels, 
to a pinion having 15 teeth, which is required to make 120 revolu- 
tions a minute, or 6 to 1 of the first wheel ; what number of teeth 
should be on each of the intermediate wheels 1 

y\/l20 X 20 = 49 mean velocity. 
72 X 20 -h 49 = 29.4 teeth on 1st driven. > A 
120 X 15 -r- 49 = 36.7 teeth on 2d driver. $ JLnS ' 
72 V 37 72 V 36 

Proof ' 15X29 6| ; ° r ' 15X^9 = 5 " 96 X 20=119 + revolutions - 
Example. — A wheel of 72 teeth, making 20 turns a minute, is to 
drive another, on whose shaft is a wheel of 36 teeth, that is to drive a 
pinion at the rate of 120 turns a minute ; how many teeth must be on 
the intermediate wheel, and how many on the pinion ? 

Vl20 X 20 = 49, and 

72 X 20 -r- 49 = 29 teeth on intermediate wheel. > A 
36 X 49 -^ 120 = 15 teeth on pinion. 5 Ans ' 

Example. — Awheel having 72 teeth, and making 20 revolutions a 
minute, is to drive another wheel, on whose shaft is a pulley of 28 
inches diameter, from which pulley motion is to be conveyed to an- 
other pulley, required to make 120 revolutions a minute ; what num- 



CIRCULAR MOTION. 317 

ber of teeth should the intermediate wheel have, and what must be 
the diameter of the last pulley 1 
V120X20 = 49, and 

72 X 20 -f- 49 = 29 teeth on intermediate wheel. > . 

28X49 -f- 120=11-13 inches diameter of pulley. \ Ans - 

The distance, from centre to centre, of two wheels to work in contact, 
given, and the ratio of velocity between them, to find their requisite 
diameters. 

Rule. — Divide the given distance by the given ratio, plus 1, and 
the quotient will be the radius of the smaller wheel ; subtract the 
radius of the smaller wheel from the given distance, and the differ- 
ence will be the radius of the larger, which, multiplied by 2, respect- 
ively, gives the required diameters of each. 

Example. — The distance from centre to centre of two shafts is 28 
inches, and one shaft is required to make three revolutions while the 
other makes one ; what must be the diameters of the wheels which 
turn the shafts, (measured from their pitch lines,) to produce the 
required effect 1 

28 -1- 4 = 7, and 28—7 = 21 ; hence, 

7 X 2 = 14 in., diameter of smaller. ) . 
21 X 2 = 42 in., diameter of larger. $ Aw5# 

Example. — The distance from centre to centre of two shafts is 40 
inches; one makes 44 turns a minute, and the other is to make 110; 
what must be the diameters of the wheels at their pitch lines? 
110 -7- 44 = 2. 5 ratio or mean velocity ; then, 



40 -4- 2.5 -j- 1 = 11.43 X 2 = 22.86 
40 — 11.43 = 28.57 X 2 = 57.14 in. 



I 



Ans. 



To find the velocity of a belt. 
The velocity of a belt is equal to the surface velocity of any drum 
or pulley over which it runs, or which it turns, in equal times and 
terms of measurements. 

Example. — A drum whose diameter is 6 feet makes 120 revolu- 
tions a minute ; what is the surface velocity of the drum per minute, 
or what is the velocity of the belt per minute 1 

6 X 3.1416 X 120 = 2262 feet. Ans. 
Velocity of belt -r- revolutions of drum = circumference of drum. 
Velocity of belt -f- circumference of drum = revolutions of drum 

To find the draft on a machine. 
General Rule. — Multiply, continuously, all the driven wheels, 
by way of their teeth, and the diameter of the front roller, together, 
27* 



318 TEETH OF WHEELS., &C. 

and, in like manner, all the drivers, by way of their teeth, and the 
diameter of the back roller, together, and divide the former product 
by the product of the latter ; the quotient is the draft. 

Example. — The driven wheels of a drawing- frame head have, 
one 72 teeth, the other 40, and the diameter of the front roller is 1.^ 
inches ; the drivers have, one 25 teeth, the other 30, and the diameter 
of the back roller is ^ of an inch ; what is the draft ? 

72 X 40 X 1.1 -J- (30 X 25 X .9) = 4.69+. Arts. 

Example. — The pinion on the front roller of a spinning frame has 
40 teeth, and the diameter of the roller is l£ inch ; the wheel on the 
back roller has 50 teeth, and the diameter of the roller is | of an inch; 
the stud gears have, driver, 21 teeth, driven, 84 ; what is the draft? 
84 X 50 X 1.5-r- (40 X 21 X .75)= 10. Ans. 

* To find the revolutions of the throstle spindle. 

Rule. — Multiply the diameter of the cylinder by the number of 
revolutions it makes in a given time, and divide the product by the 
diameter of the whir ; the quotient will be the number of revolutions 
of the spindle made in the same time. 

Example. — The diameter of the cylinder is 8 inches, and makes 
480 revolutions a minute ; the diameter of the whir is # of an inch ; 
how many revolutions are made by the spindle per minute ? 
480 X 8 -r- .875 == 4388.6 revolutions. Ans. 

To find the number of twists per inch given to the yarn by the throstle 
Rule. — Divide the number of revolutions of the spindle, in any 
given time, by the number of revolutions of the delivery (front) 
roller, multiplied by its circumference, in inches, made in the same 
time. 

Example. — The diameter of the front roller is f of an inch, and 
makes 110 revolutions a minute ; the spindle revolves 4388.6 times in 
a minute ; what number of twists per inch has the yarn ? 

.75 X 3.1416 = 2.3562 inches circ. of roller; then, 
4388.6-^(110 X 2.3562) =17 twists, nearly. Ans. 



EXPLANATIONS. 

Pitch Line. —A circle defining the base of the working or impinging section of the 
teeth. 

Pitch of a Wheel. —The distance from centre to centre of two adjacent teeth, meas- 
ured upon their pitch line. 

Length of a Tooth. — The distance from its base to its extremity. 

Breadth of a Tooth. — The length of the face of the wheel. 

Thickness of a Tooth. — The chord of the arc described upon it by the pitch line , the 
greatest cross section to the breadth. 



TEETH OF WHEELS, ETC. 319 

Circumference of a Wheel. — The pitch line, and its diameter is measured therefrom. 
See diagram — Geometry. 

Pitch X 2.5 = breadth. J Pitch X -47 == thickness. 

Thickness X 15348 = length. Length X .65154 = thickness. 

Thickness X 2.1277 = pitch. \ Thickness X 5.31925 = breadth. 

Pitch X number of teeth X -313 = diameter. 

Diameter -£- pitch X -318 = number of teeth. 

Diameter -f- number of teeth X -318 c= pitch. 

As the number of teeth on the wheel, -j- 2.25, are to the diameter 
of the wheel, so are the number of teeth or leaves on the pinion, 
-f- 1.5, to the diameter of the pinion. 

Example. — A wheel, 16 inches in diameter, and having- 81 teeth, 
is to pitch with a pinion having 25 leaves ; what must be the diameter 
of the pinion'? 

81 + 2.25 : 16 :: 25 -f- 1.5 : 5.093 inches. Ans. 

As the number of teeth on the wheel, -j- 2.25, are to the diameter 
of the wheel, so are half the number of teeth on the wheel, -f- half 
the number of leaves on the pinion, to the distance their centres should 
have. 

Example. — A wheel is 16 inches in diameter, and has 81 teeth ; 
the pinion with which it is to work has 25 leaves ; what should be 
the distance from the centre of the wheel to the centre of the pinion? 
81 + 25 = 106 -T- 2 = 53 ; then, 
81 -f- 2.25 : 16 :: 53 : 10.1862 inches. Ans. 
As half the number of teeth on the wheel and pinion are to the dis- 
tance from centre to centre of the wheel and pinion, so are the num- 
ber of teeth on the wheel, -j- 2.25, to its diameter ; or, so are the 
number of teeth on the pinion, -j- 1.5, to its diameter. 
53 : 10.1862 :: 83.25 : 16 inches. 
53 : 10.1862 :: 26.5 : 5.093 inches. 

As the number of teeth on the wheel and pinion, — 3.75, are to the 
distance between the centres of the wheel and pinion, so is the cir- 
cumference of the wheel to the diameter of the pinion, very nearly, 
and proving the almost strict accuracy of the foregoing. 

106 — 3.75 : 10.1862 :: 16 X 3.1416 : 5.0075 inches. 

To find the horse power, at a given velocity, of a cast iron tooth con- 
structed on the foregoing principles. 
Rule. — Multiply the breadth of the tooth, in inches, by the 
square of its thickness, in inches, and divide the product by twice the 
length, in inches ; the quotient, multiplied by the velocity in feet per 
second, gives the reliable horse power at the velocity specified. 

Example. — The teeth on a wheel have each a breadth of 10.64 
inches, a thickness of 2 inches, and a length of 3.07 inches, 



320 HYDROSTATICS. 

required their reliable strength, in horse power, at a velocity of 6 
feet per second. 

10.64 X 2 2 X 6 -7- 3.07 X 2 = 41.59 horse power. Ans. 

JOURNALS OF SHAFTS. 

To find the requisite diameter of a cast iron journal to resist torsion and 
stress in overcoming a given resistance at a given velocity. 

Mr. Buchanan gives deductions, from which are derived the following 
rule, for ascertaining the proper diameter of the journal of a water- 
wheel, or first mover, in any machine. 

Rule. — Multiply the resistance, in horse power, by 400, and 
divide the product by the number of revolutions of the wheel per 
minute ; the cube root of the quotient is the requisite diameter of the 
journal, in inches. For shafts inside of the mill, to drive smaller 
machinery, use 200, instead of 400, for the multiplier, and, if the 
shafts are to drive still smaller machinery, use 100 as the multiplier. 

Mr. Grier gives the mean of all these, or 240, as the multiplier, to 
resist torsion alone, and directs to take, for second movers, the diam- 
eter thus found, multiplied by 0.8, and for third movers the same 
diameter multiplied by .793. 

If the journal is wrought iron, multiply the diameter, found by the 
preceding rule, by .963 ; if of oak, by 2.238. 

Example. — The diameter of a water-wheel is 16 feet ; the resist- 
ance it has to overcome (at its pitch with the jack) is 40 horse power, 
and the surface velocity of the wheel is 6 feet per second ; what should 
be the diameter of its journals ? 

60 X 6 -j- 16 X 3.1416 = 7 revolutions of wheel per minute ; and 
40 X 400 -T- 7 = ^/2286 » 13.2 inches. Ans. 



HYDROSTATICS. 

All fluids, at rest, press equally in every direction. The pressure 
exerted by them, therefore, can never be so little as their weight, and 
may, under circumstances, be to almost any conceivable extent 
greater. The downward pressure exerted by a fluid is its weight, 
and its weight is as the quantity ; but the lateral pressure exerted is 
in a measure independent of quantity, being dependent upon depth, or 
vertical height. 

Any given area, in any given section of a containing vessel, is 
pressed equal to the weight of a column of the fluid whose base is 
equal to the area pressed, and whose height is equal to the distance 
of the centre of gravity of that area, below the surface of the fluid ; 



HYDROSTATICS. 321 

this is the case whether the sustaining surface be horizontal, or ver- 
tical, or oblique. 

The bottom of a containing vessel, therefore, whatever be its shape, 
sustains a pressure equal to the weight of the superincumbent fluid, 
or equal to the weight of a column of the fluid whose base is equal to 
the area of the bottom, and height equal to the distance from the bot- 
tom to the surface — equal to the area of the bottom, multiplied by 
the depth of the liquid, multiplied by its weight, in like terms of 
measurement. 

And each side of the containing vessel, whatever number of sides 
there be, sustains a pressure equal to the area of that side multiplied 
by half the depth of the liquid, multiplied by its weight, in the same 
terms of measurement. 

Thus, a rectangular vessel, whose sides and bottom are equal, and 
each two feet square, has a capacity of 8 cubic feet ; it will hold, 
consequently, 8 cubic feet of fresh water, one cubic foot of which 
weighs 62£ lbs. It will hold, therefore, 62£ X 8 = 500 lbs. of water. 
Now, if we suppose this vessel filled with water, we have, according 
to the foregoing, a pressure on the bottom of 2 X 2 X 2 X 62.5 = 
500 lbs. ; a pressure exactly equal to the weight of all the fluid. And 
we have, upon each of the four sides, a pressure of 2 X 2 X 1 X 
62.5 = 250 lbs. ; a lateral pressure, therefore, equal to 250 X 4 = 
1000 lbs., — equal to twice the pressure on the bottom, and showing 
the entire pressure exerted to be 300 per cent, greater than the weight 
of the water employed. 

Again : if we suppose the above vessel contracted, laterally, to the 
extent that its sides are but 3 inches, or \ of a foot apart, through- 
out, and that its length is so extended that it still holds the 8 cubic 
feet of water, then we have, upon the bottom, whose area is only 9 
square inches, a pressure of .25 X .25 X 128 X 62.5 = 500 lbs. 
as before ; and upon each side we have a pressure of .25 X 128 X 
.128 x 62.5 = 128000 lbs. ; making in all a pressure of 128000 X 
4 -{- 500, — the enormous pressure of 512500 lbs. ; and that too ex- 
erted by 8 cubic feet or 500 lbs. of water. It is easy to see that the 
same principles hold good under any extent of lateral area. 

Example. — A sluice or flood-gate is 3 feet by 2£, and its centre is 
12 feet below the surface of the water ; what pressure does the water 
exert upon it % 

3 X 2.5 X 12 X 62.5 = 5625 lbs. Ans. 

Example. — A dam, that presents a perpendicular resistance to a 
stream, is 40 feet long and 15 feet high ; the water is level with its 
top ; what pressure does the dam sustain, supposing the water at rest, 
and what is the mean pressure against it per square foot? 
40 X 15 X ^ X 62.5 = 281250 lbs., pressure against the dam; and 
281250 -£- 40~X~15 = 468| lbs., mean pressure per sq. foot. Ans. 



322 HYDRAULICS. 

Example. — The same stream, the same length of dam, and the 
same vertical height as the preceding, and the dam sloping into the 
stream against the current, 30 feet from its base ; required the pres- 
sure against the dam, and the average pressure per square foot. 
40 X 15 X 7.5 X 62.5 = 281250 lbs., press, as before. -v 

\/l5 2 -{- 30 2 = 33.5jU feet, slant height of dam ; and ( Ans. 

281250 -r- 40 X 33.541 = 209.63 lbs. av'g pres. per sq. foot. ) 



HYDRAULICS. 

The established law for the velocity of all bodies falling from rest 
is given under Gravitation, viz., that V height X 64.33, or A/height 
X 8.02 = velocity per second, or velocity in one second of time, the 
velocity and height both being in the same denomination of measure. 
And from what has been said concerning pressure, under Hydro- 
statics, it is evident that the same law will cause water, or other 
fluid, to flow through an opening in the side of a reservoir, or dam, 
with the same velocity that a body would attain falling perpendicu- 
larly through a space equal to that between the surface of the water 
and the centre of the opening alluded to ; and that, consequently, the- 
oretically, the quantity thus discharged, in any given time, will be 
equal to the product of the velocity and area of the opening, multi- 
plied by that time. 

The theoretical law, however, last adduced, under ordinary circum 
stances, does not apply. And the quantity discharged, owing to the 
contraction of the fluid vein, caused by the friction of the particles 
against the sides of the opening, falls short of that theoretically due. 
The only instance known in which the full force of the law may be 
obtained, is where the discharge is made to issue through a straight 
tube whose form is the frustum of a cone, its length being half the 
diameter of the aperture, and the diameter of the receiving end to 
that of the discharging end as 5 to 8 ; when a fluid is allowed to pass 
through such an opening, no contraction of the vein takes place. 

From various carefully conducted experiments by M. Morin, Eytel- 
wein, Bossut, and others, the following practical rules for ascertain- 
ing the quantity discharged through different openings, and under dif- 
ferent heads, are derived : — 

1. When the issue is through a circular opening, its upper vertical 
point as high as the surface of the fluid, estimate the height or head 
from the centre of the opening to the surface of the fluid, and use 5.4, 
instead of 8.02, as the coefficient of quantity. 

2. When the opening is circular, and under a head equal to its 
diameter, estimate the head as in the preceding, and use 7| as the 
coefficient. 



WATER-WHEELS. 323 

3. When the issue is through a rectangular orifice, two or more 
feet beneath the surface, estimate the head from the centre of the ori- 
fice to the surface of the water, and use 5.1 as the coefficient. 

4. When the discharge is from a rectangular opening, extending as 
high as the surface of the fluid, estimate the head from the bottom of 
the opening to the surface of the water, and use 3.4 as the coefficient. 
This rule applies to water flowing over a dam, or from a notch or slit 
cut in its side, &c. 

It may be proper to add, that if the orifice is small and under con- 
siderable head, the quantity discharged, relatively, will be slightly 
less than would be discharged if the opening were nearer the surface. 

From the foregoing we obtain the following 

GENERAL RULE. 

Multiply the square root of the height, or head, (as estimated in 
the foregoing,) in feet, by the coefficient of quantity given as per- 
taining thereto, and the quotient will be the effective velocity in feet 
per second of the discharge ; which, multiplied by the area of the 
opening in feet, gives the quantity in cubic feet discharged in a single 
second, or in each second of time. 

Example. — A rectangular opening in the side of a dam is 6 feet 
long and 8 inches deep ; and the distance from the centre of the open- 
ing to the surface of the water is 4 feet ; required the quantity of 
water discharged in each second of time. 

V4 = 2X5. 1X6X1 = 40.8 cubic feet. Arts. 

Example. — A dam is 60 feet long, and the water flows over its 
entire length 6 inches, or £ foot deep ; what quantity flows over per 
second ? 

V.50* = .7071 X 3.4 X 60 X -5 = 72 1 cubic feet. Ans. 



WATER-WHEELS. 

The many uncertainties and doubts which existed until lately con- 
cerning the best mode of constructing a water-wheel, with the view 
to obtain a maximum of effect, the velocity at which the wheel should 
move, and other requisites pertaining to it generally, have, in a great 
measure, through investigations and experiments by the Franklin In- 
stitute in this country, added to those by M. Morin, in France, and 
other parties interested, been removed ; and the whole seems now to 
have nearly subsided into the following general and demonstrative 
conclusions : — 

* The decimal .50 is the same value as .5, = i, but it will be recollected that to obtain 
the root of a decimal full periods must be used. 



324 WATER- WHEELS. 

1. That, to obtain a maximum of effect by a horizontal water- 
wheel, the water must be laid upon the wheel on the stream side, and 
at a point or line on the wheel about 52| degrees distant from its sum- 
mit; or, the effective fall — distance from the centre of the discharg- 
ing orifice to the bottom of the wheel pit, or water therein, when the 
wheel is at rest — being 1, the diameter of the wheel should be 
1.108. 

2. That the periphery of the wheel ought to move at a velocity in 
feet per second equal to about twice the square root of the number 
of feet effective fall ;* that the number of buckets should equal 2.1 
times the wheel's diameter in feet ; and that due means be adopted for 
the escape of the air from the buckets, either by causing the stream 
to flow some inches narrower than the wheel, or otherwise. 

3. That a head of water is requisite sufficient to cause the velocity 
pf its flow to be as 3 to 2 of the velocity of the wheel (ordinarily 
not less than two nor more than three feet). 

4. That a wheel of good workmanship, - constructed and geared 
according to these restrictions, will return, as a maximum, about 80 
per cent, of the power employed. 

5. That, because of water producing a less effective power by im- 
pulse than gravity, turbines, or wheels through which the motion is 
obtained by reaction, are greatly preferable to undershot or low-breast 
wheels ; that they are, seemingly, as well adapted to great as to small 
falls, returning, in either instance, under favorable circumstances, an 
useful effect of from 70 to 78 per cent, of the power expended ; that 
their velocities may vary considerably from that affording the maximum 
effect; (§ of their light velocity,) without materially diminishing their 
effect; that they are nearly as effective when drowned to the depth of 
several feet as when working free, thereby making use of a greater 
fall than can be obtained, at the same locality, for any other wheel ; 
and that they receive variable quantities of water without altering 
the ratio of the power to the effect. 

These considerations, taken in connection with the less important, 
that they are durable, not more liable than others to require repairs, 
occupy less room in their position, and cost decidedly less than other 
motors of equal efficacy, are fast bringing this class of wheels into 
favor and use. 

To find the power of a stream to turn an overshot or pitched-back wheel. 
Example. — The entire height — head and fall — of a stream is 19 feet, and the quan- 
tity of water flowing is that which may be drawn through a rectangular opening 16 feet 
long and 3 inches deep ; required the greatest exertive force of the stream, in horse- 
power, applied as above supposed. 

16 X 12 X 3-r 144 = 4 feet area of discharge. 
19 X -1221 = 2.32 feet requisite head. 
^2.32 = 1.523 X 5.1 = 7.76 feet, effective velocity per second. 

♦ The practice is becoming very general to gear all wheels, great or small, to a velocity 
of about 6 feet per second. 



WATER-WHEELS. 



325 



7.76 X 4 = 31 cubic feet discharged per second. 
31 V 60 = 1860 cubic feet discharged per minute. 
1860 X 62.5 = 116250 lbs. discharged per minute. 



116250 X 19 — 2.32 = 1939050 lbs. momentum. 
1939050 + 33000 = 53$ horses' power. Ans. 

To find the requisite dimensions of a wheel, based upon the preceding principles, 
adapted to the foregoing stream. 

19 — 2.32 = 16.68 feet effective fall. 

16.68 X 1-103 = 18.48 feet diameter of wheel. 

360° : 18.48 : : 52°. 75 : 2.7 feet of wheel above discharging orifice. 

18.48 — 2.7 = 15.78 feet of wheel below discharging orifice. 

16.68 — 15.78 = 0.9 foot clearance of wheel. 

V 16.68 = 4.08 + X 2 = 8- 16 feet velocity of wheel per second. 

18.48X2.1 =39 buckets. 

18.48 X 3.1416 = 58.06 feet circumference of wheel. 

8.16+ X 60 = 490.17 feet velocity of wheel per minute. 

490 -J- 53 = 8.44 revolutions of wheel per minute. 

1860 -j- 490 = 3.8 feet sectional area of buckets. The buckets, to properly retain tho 
water and avoid waste, should be but half full ; therefore — 

3.8 X 2 = 7.6 feet practical sectional area ; and to allow sufficient room for the escape 
of the air, the wheel should be, say 9 feet broad. 

7.6 -7- 9 = .85 — , say 1 foot depth of shrouding. 

I860 — 39 X 8.44 = 5.68 cubic feet of water received by each bucket. 

68 -7- 39 = 1.49 foot breadth of bucket. 

5.68 — 1.49 = 3.8 feet sectional dimensions, as before. 

3.8 X 2 X 1-49 = 11.36 feet practical capacity of buckets, more, allowance as above. 

From a well constructed wheel the water begins to empty at about 5 feet from the bot- 
tom; ther efore — 

18.48 — 2.32 + . 5 + 5 = 10.66, ratio of diameter to loaded arch. 

18.48 : -3 9 : : 10.66 : 11.2 loaded buckets. 

11.2 X 5.68 X 62.5 = 3971 lbs. on loaded arch. 

Or, a very good and safe rule for determining the weight, in pounds, constantly on tho 
wheel, is this : — Multiply |j- of the buckets on the wheel by the number of cubic feet of 
water received by each, and that product by 40. Ex. 

39 + 9 = 4J X 4 X 5 - 6 3 X 40 = 3938 lbs. load. 
3971 X 490 = 1945790 lbs. ; or, ) momentum, or 

39- X 8.44 +X 5.68- X 62.5 X 16.68 = 1939050 lbs. \ exertive force. 
1939000 + 33000 = 58.75 X -75 = 44 h. p. effective. Ans. 

From the foregoing a very good practical rule is derived for determining the requisite 
head for an y given veloci ty of wheel ; thus — 

height -f- 8.16 : 2.32 : : height + required velocity : required head. 

Example. — The entire height is 16 feet, and the velocity of the wheel is to be 6 feet 
per second ; required the necessary head. 

16 + 8.16 : 2.32 : : 16 + 6: 2.11 feet. Ans. 

It has been demonstrated that in practice nearly two feet head is required to generate a 
velocity of 5 feet per second, and this rule gives 

10+ 8.16: 2.32 : : 10 + 5 : 1.916; or, 25 + 8.16 : 2.32 : : 25T5 : 2.099, showing a 
difference of only .18, or a mean error of about one inch, between the two extremes — 10 
feet head and fall, and 25 feet head and fall. 

If it were intended to construct a wheel for the foregoing stream, to run at a less veloc- 
ity, say at 6 feet per second, (the other restrictions to be maintained,) then we should 
have 

6 X 60 = 360 feet velocity of wheel per minute. 

360 + 58.83 = 6.12 revolutions of wheel per minute. 

I860 + 360 = 5 feet sectional area of buckets. The depth of the buckets should sel- 
dom vary much from 12 inches ; the required breadth of the wheel, therefore, in this 
instance, would be about 11 feet. But it is customary, perhaps unadvisedly, when the 
Velocity is intended at 6 feet or less, to place the buckets a little nearer together than com* 
28 



326 DYNAMICS. 

ports with the preceding; in which case, of course, a less breadth of wheel 
would be required. 

If we multiply the number of cubic feet flowing upon the wheel per minute, 
by the effective fall, and divide the product by 700, we obtain the effective 
horse-power; thus — 

1860 X 16.68 -h 700 = 44.3 horses' power. 

The power should be taken from the wheel on the side to which the water is 
applied, and at a point horizontal to the centre of the wheel; and belts, in- 
stead of gears, for all the prime movers, should be used as far as practicable. 



DYNAMICS. 

UNIFORM MOTION. 

P =z mechanical power in pounds, or power in effects. 

F = force in pounds, or weight or resistance to be overcome. 

V = velocity of the force F in feet per second. 

t = time in seconds during which F is in motion. 

s = space in feet through which F moves in the time t. 
m = momentum of the force F in the time t. 
M = mass, or moving matter in effects. 
W = work in foot-pounds of power. 

Pz=FV—Fs + t. F=P+V=Pt + s. V=P+F=s-S-t. 
t = s-±-V=Fs + P. 
s=Vt = Pt + F. m = Ft = MV. M=Ft+V=W+V 2 . 
W= Fs = Ftv = M V 2 = PL 

UNIFORMLY ACCELERATED MOTION. 

F=V+gt = 2s-i-g?= V 2 + 2gs. V— 2s-i-t = Fgt=\/(2gFs). 
s = \tv = %gF? = V*-±2gF. t = 2s+V=V+gF=^(s + lgF). 

MOTION OVER A FIXED PULLEY. 

Let A and a represent the opposing forces or weights. 

A-\-a 2(A-\-d) ** ' 

g=z 32.166 feet per second. (See p. 85.) 

HYDROSTATIC PRESS. 

A = transverse area of cylinder piston. 
D = diameter of cylinder piston. 

a z=z transverse area of forcing-pump piston. 

d = diameter of forcing-pump piston. 

b = diameter of safety-valve. 

f= pressure in pounds on safety-valve that prevents it from 
rising. 

a:A:;F:P. d* ; D 2 : : F i !£\ WiVuPif. 



APPLICATIONS OF PRINCIPLES IN DYNAMICS. 327 



GENERAL APPLICATIONS OF PRINCIPLES 
IN DYNAMICS. 

To find the power of a stream issuing from an opening in the side 
of a reservoir, or flume. 

y = head of water, in feet. See Hydraulics, p. 322. 

h === effective fall, or height of the column of water, in feet. 

k = practical substitute for \/(2g) due to the form and position 
of the opening. See Hydraulics, pp. 322, 323. 

V = ksjh' = velocity of the water in feet per second through 
the opening, or discharging orifice. 

A = area of the opening, in square feet. 

Q = AV= quantity of water, in cubic feet, discharged per 
second. 
H === actual or theoretical horse-power of the stream. 
F = 62.5 Ah = weight of the column of water, in pounds. 
W = 6QFV = work in foot-pounds of power per minute. 
H= -JL = ^X62.5 X 60 = 0>1186Q k 
33,000 33,000 

To find the power of a bucket-wheel. 
n = number of buckets on the wheel. 
q = JL capacity of a bucket, in cubic feet. 

V = proposed velocity of the circumference of the wheel, in 
feet, per second. See Water-wheels, p. 324. 

F= ??^£2 X 0.8936 =15522 = weight of water, in pounds, 

7T 9 

that the wheel can constantly carry. 

W=6QFV. H= W =0.3232gnF. 
33,000 a 

To find the power of a turbine wheel. 

h 33 height, in feet, of the column of water resting on the wheel. 

A = area of the ventage of the wheel, in square feet. 

V = \f(2g)^h = S.02\/h = velocity, in feet, per second, at 
which the water enters the wheel. 

Q = AV= A\/(2gh) = quantity of water, in cubic feet, dis- 
charged per second. 

F = 62.5Ah. W=60FV. #=_JL = 0.1136Qft. 

33,000 

To calculate the power of a hydraulic ram. 
F = 62.5Ah. W=60FV. H= W -± 33,000= 0.1136QA. 
In the preceding calculations I have thought proper to express 



328 APPLICATIONS OF PRINCIPLES IN DYNAMICS. 

the actual power, force, and velocity, in all cases, and to avoid any 
allusion to what is called the effective. The effective power of a 
machine depends upon so many conditions, — the unavoidable fric- 
tion of its parts, varying more or less in different machines, sup- 
posed to be in all respects equal ; the mechanical adjustments and 
adaptations of the parts ; the material, or materials, of which it is 
composed, and the workmanship generally, — that it admits of no 
fixed mathematical formula, correctly expressive of its worth, un- 
less the working or sensible force is known ; of which I am about 
to incidentally speak, and which I shall define. 

Query in Dynamics. — Is there a quantity, which, if taken 
as a constant, will equalize force with time and space ? 

Now, to the mere mathematician this proposes nothing ; but to 
science it propounds an important question, which, I believe, has 
not been answered, if ever started before. But science answers 
nothing per se. She does, however, suggest, and seems to whisper 
that there is such a quantity. Then, if so, what is it ? 

Science : We have seen by the law of uniformly accelerative 
velocity, or gravitation, that V= \/(2g)^s = gt for any length of 
time, in which V represents the velocity at the close of the time, 
g the constant acceleratrix, s the space described in the time, and 
t the time in units of time. 

Mathematics : Wherefore V = g, and s = %g = \V in the 
first unit of time. 

Science : But \g now represents a space described by a uni- 
formly accelerative velocity. 

Mathematics : The mean of which is equal to \](\g) = 
\/(iV). 

Logic : And, therefore, equal to the uniform velocity, which is 
equal to the space, in the first unit of time. 

Science: But, in uniform velocity, spaces and velocities are 
equal to each other in all equal times. 

Logic : Therefore ^Qg) represents the space in uniform veloci- 
ty, under the influence of gravity, that is equal to each unit of 
time ; and, consequently, since the velocity divided by the product 
of the space and time is equal to the force in all cases, that is equal 
to a unit of force in each unit of time. 

Thus, there being no fallacy in the preceding, we have answered 
the first query by showing that there positively is such a quantity, 
and the second, by showing its algebraic value. The law of uni- 
form velocity, therefore, and the law of mechanical force and 
velocity, are involved in, and derivable from, the law of gravi- 
tation. 

The position taken, then, is this, viz., that one unit of mechani- 
cal force, constantly applied, will move a body equal in uniform 



HEAT — CALORIC. 329 

velocity to \^(jg) = 4.01 feet a second, the influence of gravity 
upon the body being with the same kind of force counterpoised ; 
or, in other words, that the addition of one unit of constantly ap- 
plied force to that of the same kind required to overcome the vis 
inertia, or resistance to motion in a machine, will move it, equal in 
uniform velocity to \/(jg) = 4.01 feet a second, and so, proportion- 
ally, for superadded force or forces. 

I will now introduce rules predicated upon this law, in place of 
the common rules, for calculating the power, &c, of steam-engines ; 
where, if the law be thought to be empirical, it can easiest be 
tested by experiments. 

A — transverse area of steam-cylinder piston, in inches. 

d = diameter of steam-cylinder piston, in inches. 

P = pressure per steam-gauge (the vacuum pressure included 
for a condensing engine). 

p = pressure per steam-gauge (the vacuum pressure included 
for a condensing engine) that is required to barely set the unload- 
ed machine in motion. 

V = velocity of piston in feet per second due to a unit of sen- 
sible or working force, = \/(^g) = 4.01. 

P — p = sensible, or working force. 

F= A(P — p). W= 60FV. H= W = d2 ( p —P) effective. 

33,000 174.6335 
Thus we have the effective horse-power of a steam-engine at 
given pressures, P and p ; and the normal horse-power of a steam- 
engine, p being practically known or assigned, and P being the 
maximum pressure that may be constantly carried with safety ; 
which last pressure may be assigned by theory, based on the thick- 
ness of the plate, the quality of the material, &c. 

P — p=f= s-^-Vt. P = s+Vt-{-p. 
Example. — What working-force, f, must be constantly applied 
to a body, in order to move it s, 25 feet, in t, 2^ seconds of time ? 
25 -f- (4.01 X 2-5) = 2.494 units of working-force. Ans. 
Example. — What total force, P, must be constantly applied 
to a body whose vis inertia, p, is 4 pounds, in order to move it s, 18 
feet, in t, 2 seconds of time ? 

18 ^- (4.01 X2) = 2.2444 -|- 4 = 6.2444 lbs. Ans. 



HEAT— CALOEIC. 

SENSIBLE, LATENT, SPECIFIC, ETC. 

Sensible heat is that description of heat in a body which is 
sensible to the touch, and which is indicated by a common ther- 
mometer. 

28* 



330 



HEAT CALORIC. 



>l 



Latent heat is that description of heat in a body which is 
insensible to the touch, and which the thermometer does not indi- 
cate ; but which, by chemical, or other forces, can be, more or less, 
set free and made sensible. 

The specific heat of a body' is the quantity of the sub- 
stance (?) heat, that it contains at a given density and tempera- 
ture. It is usually, however, expressed in comparison with that of 
water as the unit, or with that of atmosphere as the unit, and for 
equal weights at like temperatures, about 60° Fahrenheit. 

Let I -4- 1 represent the sum of the latent and sensible heat of 
saturated steam, at t° of sensible heat, under a pressure in units of 
force, p, of 1 atmosphere ; then 

Z+J _ H78.6 _ 2.493062, the specific heat of saturated 
pg 472.752 r 

steam, at t = 212° F., under a pressure of 1 atmosphere, p, = 14.7 
pounds per square inch. 

— + 1 = 14.9214213, the atomic weight of saturated steam 
W 
at t = 212° F., under a pressure of 1 atmosphere. 

— ±L X ( -v — ~T~ 1 J — 37.2, a constant, equal to the product of 
P9 V^ / 
the specific heat and atomic weight of all substances at t = 212°F. : 
whence 37.2 — 1 = 36.2, the atomic weight of the atmosphere; 
and 2.493062 -|- 1 -f- (9 -4- 1) = 0.3493, the specific heat of steam 
under a pressure of 1 atmosphere, water as unity. 

And, since the specific heat of the atmosphere is known to be 
0.2669 when that of an equal weight of water is 1, both at 60°, F., 
we have 

(9-4-1) -f- 0.2669 — 37.4672, a constant, equal to the product 
of the specific heat and atomic weight of all substances, at t = 
60°, F. 

pt — [y/(^) + 1] = 772.1471, the mechanical equivalent 

of a unit of heat in foot-pounds ; i.e., one unit of heat will raise 1 
pound weight 772.147 feet high. 

Now, in regard to the numerical values assigned to the elements 
introduced in the foregoing, no warranty, of course, can be offered 
that they are strictly correct ; since, at best, they are mostly but 
settled conclusions, based on the results of carefully conducted ex- 
periments. But the theory introduced, it is believed, will be found 
tenable ; nevertheless, it is offered, open to further and more criti- 
cal investigation. 

The specific heats of the gases increase as their densities decrease, 
but somewhat less rapidly than the diminutions in density ; the 
exact ratio, I believe, is not known. 



HEAT CALORIC. 



331 



The sensible beat of water is augmented by violent agitation, 
according to Mr. Dalton, as follows ; viz., from 60° F., in ± hour, 
10°; in 1 hour, 14.5°; in 2 hours, 19.5°; in 3 hours, 29.5°; in 5 
hours, 39.5° ; and in 6 hours, 42.5°. 

The sensible heat of atmospheric air is increased by sudden com- 
pression to -^ its normal bulk about 500°. 

The sensible heat of soft iron is increased by receiving a few 
rapid and strong blows with a common hammer to a bright red 
heat in day-light, equal to about 900°, and always, as in the case 
of water, more by the first blow than by the next succeeding, and 
so on. 





Sensible and latent heat 


of steam. — Regnault. 




Sensible . 


Latent °. 


Sensible . 


Latent °. 


Sensible . 


Latent °. 


Sensible . 


Latent °. 


32 
104 
140 


1092.6 
1042.2 
1017.0 


176 
212 
230 


991.8 
966.6 
952.2 


248 
266 
302 


936.6 
927.0 
901.8 


338 
374 

410 


874.8 

849.6 
822.6 



Latent heat of different substances — equal weights. 



Alcohol, 364° 

Ammonia, 860° 

Beeswax, 175° 

Bismuth, 22° 

Ether, 163° 

Specific heat of equal 
reported) 
1. 
0.2669 
0.7 
0.054 
0.023 
0.202 
0.219 



Ice, 142.6° 

Lead, 162° 

Mercury, 157° 
Phosphorus, 9° 
Spermaceti, 148° 



Steam, 

Sulphur, 

Tin, 

Water, 

Zinc, 



966.6° 

17° 
500° 
966.6° 
493° 



Water, 

Air, 

Alcohol, 

Antimony, 

Bismuth, 

Brick, clay, 

Carb. acid, 

Carb. oxid gas, 0.288 
0.13 
0.241 
0.15 
0.202 
0.095 
0.517 
0.195 
0.421 
029 
032 



weights of different substances (as commonly 
compared with that of water as 1. 



Cast iron, 
Chareqal, 
Cobalt, 
Coke, 
Copper, 
Ether, 

Glass, crystal, 
Gas of oils, 

Gold, 



V. 

{o. 



Hydrogen, 

Ice, 

Iron, wrought, 

Lead, 



tt- 



Lime, 

Linseed oil, 
Mercury, 
Nickel, 
Nitric acid, 
Nitrogen, 

Nitro-oxid gas, 0.231 
0.31 
218 



3.405 
3.294 
0.504 
0.113 
031 
029 
0.217 
0.53 
0.032 
0.104 
0.661 
0.275 



Olive oil, 

Oxygen, 

Petroleum, 
Phosphorus, 



ft 



361 
0.468 
0.189 



Platinum, 0.034 

Silver, 0.057 

( 0.475 
Steam, \ 0.347 

(0.365 

Steel, 0.116 

Spts. turpentine, 0.467 

c i -u ( 0.209 

Sulphur, | 0<188 

Sulphuric acid, 0.334 

Tin I " 048 

xm > 1 0.056 

Tellurium, 0.091 

Woods, avg., 0.51 

Zinc, 0.095 



332 



HEAT CALORIC. 



Capacity for heat expresses the relative ability of a body to re- 
tain its respective specific heat, compared, usually, with that of 
water as the unit, and at equal weights, or equal volumes ; and the 
capacities for heat of different bodies are inversely as their den- 
sities. 

Relative capacities for heat of different substances (as commonly 
reported}. 



Names of 


Equal 


Equal 


Names of 


Equal 


Equal 


bodies. 


weights. 


volumes. 


bodies. 


weights. 


volumes. 


Water, 


1.000 


1.000 


Iron, 


0.126 


0.993 


Brass, 


0.116 


0.971 


Lead, 


0.043 


0.487 


Copper, 


0.114 


1.027 


Mercury, 


0.036 


.... 


Glass, 


0.187 


0.443 


Silver, 


0.082 


0.883 


Gold, 


0.050 


0.966 


Tin, 


0.060 


.... 


Ice, 


0.900 




Zinc, 


0.102 






SECTION VI. 

COVERINGS OF SOLIDS OR PROBLEMS IN 
PATTERN CUTTING. 



Under this head, I propose chiefly to contemplate patterns that 
are applicable to the wants and purposes of Tin-Plate and 
Sheet-Iron Workers ; and, in treating of the construction of these, 
my main purpose will be to offer clear and unmistakable step-by- 
step directions for constructing them practically, and, as far as ad- 
missible by mechanical means ; to the end that the student unac- 
quainted with the principles involved in his tasks, and reluctant to 
enter into mathematical calculations, can nevertheless accomplish 
his purposes, and with accuracy and despatch : and I shall accom- 
pany the proceedings with diagrams for illustration and reference. 
But since I shall be obliged to view the patterns theoretically and 
analytically in all their parts, in order to devise the best rules for 
constructing them practically, I shall deem it not unadvisable, with 
reference to many of them at least, to state the laws and the math- 
ematical data upon which the directions for their construction will 
be predicated. Moreover, many of the patterns will be found of 
a high geometrical type, governed by inflexible laws, and capable 
of mathematical investigation and measurement in all respects ; 
and to present these only in their naked aspects of mere mechani- 
cal contrivances, to be fixed in the memory, or copied at will, 
would seem out of place in a work of this kind. 

Problem 1 of the following series will embrace in its solution 
all the principles involved in the construction of the whole class of 
patterns to which it will relate ; and, more or less interwoven with 
its solution mechanically, I shall, once for all, with reference to 
the class, enunciate the laws and define their bearings, to the full 
end of constructing them theoretically and mathematically ; and I 
shall do this in as brief a manner and as free from technicalities as 
the circumstances will admit of. 

But, before proceeding to the solution of problems, it may be 
proper, perhaps, to explain the meaning of some terms that I shall 



334 PATTERN CUTTING. 

be liable to make use of; and it may as well be done here, per- 
haps, as elsewhere. 

A vessel in the form of a frustum of a cone or truncated cone, 
is a flaring vessel having circles for its bases. 

The lateral portion of a conical vessel or cylindrical vessel is 
the side or body. 

The bases of a vessel are the ends. 

The fixed base of a vessel is called the bottom ; and the mova- 
ble base, the cover. 

The slant depth of a vessel (and none but flaring vessels have 
slant depths) is the depth of the side. 

The diameters of a vessel are the diameters of the bases or 
ends. 

The perpendicular depth of a vessel is its depth. 

The circumference of a circle is equal to the diameter multi- 
plied by 3.1416 ; or it is equal to the diameter multiplied by 355, 
and divided by 1 1 3 ; or it is nearly equal to the diameter multi- 
plied by 22, and divided by 7. 

The diameter of a circle, therefore, is equal to the circumfer- 
ence divided by 3.1416 ; or it is equal to the circumference mul- 
tiplied by 113, and divided by 355 ; or it is nearly equal to the 
circumference multiplied by 7, and divided by 22. The Greek let- 
ter 7r, if met with in connection with the problems, will invariably 
mean 3.1416, or the ratio of the circumference to the diameter, 
the latter being 1. 

= Solidity ; A = Area ; & — Curved or convex surface of a 
Solid. 

In geometry, written lines are limited by the letters or charac- 
ters that are placed at their extremities ; and, in the text, they are 
announced by the same letters or characters written with a space 
between them. Thus a b, k m, c z, Sfc, in the text refer to the lines 
limited by a b, k m, c z, fyc, in the diagrams ; but the values of these 
lines, that is their lengths, when they are introduced into equa- 
tions by the letters that limit them, are otherwise expressed : thus 
a & or (a b) in the text or an equation means the length of the line 
a &, or that is limited by a and b ; ab" or (a &) 2 means the square of 
the length or line a b ; 2 a b or 2 (a b), twice the length of the line 
a &, fyc. 

In algebraic notations, factors and numeral co-efficients and 
factors are usually written without the sign of multiplication or 
a space between them; thus abc, 2a, I2ab, (ab — d)(cd-\-m), 
are to be read, a X b X c, 2 X a, 12 X a X b, (a X b — d) 
(c X d-\-m), that is, the difference of d and the product of a into b 
is to be multiplied by the sum of the product of c into d and the 
quantity m. 



PATTERN CUTTING. 



335 



Prob. 1. — To construct a Pattern for the Lateral Portion of a ves- 
sel in the form of a Frustum of a Cone of given Diameters and 
Depth. 

The chief principle involved in the construction of this descrip- 
tion of patterns is easily explained : it is that of a right cone 
placed upon its side, and rotating on a plane. If a cone so placed 
and starting from rest make one revolution, or, in other words, roll 
once over, its whole lateral surface, correctly delineated, may be 
supposed to be described upon the plane ; if it make a half-revolu- 
tion or roll half-over, half its lateral surface in like manner delin- 
eated may be supposed to be described ; and so on for any partial 
or fractional rotation whatever : thus the slant height of the rotat- 
ing cone will be the radius of the arc that will be described by the 
rotating base, and the arc so described will be that of the lateral 
surface, lateral portion, side, body, or covering. The rotating 
cone, then, that will describe the greater arc of the lateral surface 
of a frustum, must be a cone including the frustum ; and that that 
will describe the lesser must be the same cone, less the frustum. 



Rule. — Place the square suitably on the plate from which the 
pattern is to be taken, as 
a m R, diagram, and draw 
to any sufficient length a 
line m R ; and from m, on 
the other arm of the square, 
set off a known measure 
(the whole or any desired 
aliquot part, as £, ^-, -|) of 
the greater given base, m a, 
and draw that measure ; 
then drop the square per- 
pendicularly down on the 
line m R, from m, equal to 
the given depth of the ves- 
sel, m h, and set off in like 
manner the same known 
measure of the lesser given 
base, h c, and draw that 
measure ; then draw a line, 
a c it, through the points 
a and c, to the line m it, 
and cutting that line in R. Next, with the distance R a in the di- 
viders or on the beam compasses, and R the centre, describe an 
arc, as a b, to any sufficient length ; and with the distance R c, and 




336 PATTERN CUTTING. 

R the centre, describe a parallel arc, as c d, to any sufficient 
length. 

We have now defined the curves for the level surfaces (top and 
bottom) of the given vessel, and the depth of its sides ; or, in other 
words, we have defined the ratio of the given diameters and the 
slant depth of the vessel ; and have thus far a pattern, in some 
sort, for a vessel of this general form, having the same slant depth, 
the same ratio of diameters, and the diameters varying from al- 
most nothing to almost twice the radii to which the respective arcs 
have been drawn ; or, in another point of view, having the same 
slant depth, the same ratio of diameters, and a depth varying from 
almost nothing to almost the depth of the side. A right section 
taken out indiscriminately, acvn, for example, would be a pat- 
tern for this kind of vessel at a fixed slant depth and a fixed ratio 
of diameters, and such patterns are sometimes used ; but clearly 
it would be no pattern for a vessel of a given perpendicular depth 
and given diameters, since it would be no known measure of the 
lateral surface required. The slant depth and the ratio of the 
diameters remaining constant, the perpendicular depth varies, as 
we have seen, with the diameters ; and the given diameters, it ap- 
pears, have as yet in no degree been fixed or defined, only their 
ratio has been defined. But since the perpendicular depth varies 
with the diameters, and the ratio of the given diameters has been 
defined, it follows, that, if we were to fix the given depth, we 
should thereby fix the given diameters ; so, if we were to fix one 
of the given diameters, we should thereby fix the given depth and 
the other given diameter. But we cannot fix the depth of a flar- 
ing vessel by describing it upon the side ; nor can we in any way 
fix the given diameters, except by their circumferences upon the arcs 
which we have drawn. 

It thus appears, that, before we can proceed to complete our pat- 
tern, we must know the circumference of one of the given bases, 
and must decide what the measuring unit of the pattern shall be ; 
whether one covering the whole lateral surface of the vessel, or 
only a right section of that surface, covering a known aliquot part, 
as |-, I-, |, or less, of that surface. 

Suppose, now, for example, that one of the given diameters of 
the pattern thus far drawn, we will say the greater, is 18 inches, 
and that we would take out a pattern covering say one-fourth part 
of the lateral surface or side ; then 3.1416 X 18 = 56.5488 inches, 
the circumference of the greater given base ; and 56.5488 -)-4 = 
14. 137, -or 141 inches nearly, the length of the greater arc that 
will contain one-fourth part of the circumference. Then, with a 
strip of flexible tin cut to that length, and bent to the curve, or by 
any other mechanical means, measure off on the greater arc from 



PATTERN CUTTING. 337 

a, in the direction n, 14§- inches large, as from atob; and from the 
new-found point, b, draw a right line to the central point, R, as 
b d R : then will the section a b d c be the pattern required. 
This may be taken out with or without the requisite margins for 
locks, burrs, &c, as desired, properly marked, and kept for future 
use. 

Note. — In order to avoid uncertainty and confusion, I shall in all cases 
in this work confine the directions to the construction of the dimensions- 
pattern, leaving the workman to allow the requisite margins for locks, 
seams, burrs, and rolled or wired rims, as his taste and the circumstances 
may require. I will suggest, however, that, for a straight lock, the allowance 
should be equal on each side, and that two-thirds of the allowance on a side, 
less the thickness of the plate, should be turned when neatness and accuracy 
of dimensions are intended ; also that, ordinarily, three-eighths of an inch ia 
sufficient for the lock on tin-plate, and one inch for that on stove-pipe sheet- 
iron. 

By Mathematics. 

R = radius, or slant height, of generating cone = R a, diagram. 

s = slant height of given frustum = c a, diagram. 
H=z perpendicular height of generating cone =R m, diagram. 
h =: perpendicular depth of given frustum = h m, diagram. 
D = diameter of base of generating cone, or of greater base of 

frustum. 
d = diameter of lesser base of frustum. 

r = radius, or slant height, of cone whose base is the lesser base 
of the given frustum = R c, diagram. 



H(D- d) Hs I . /D — dY „ I ; A*Y 

_ Hd Rd ds 

D- w — l = — = -- + a=tyR>-H° = tys>-hS> + d, 

= D—2Vs°- — h\ 
29 



338 



PATTERN CUTTING. 



Pros. 2. — To construct a Pattern for the Body of a Vessel in the 
form of a Frustum of a Cone of given Dimensions, without plot- 
ting the dimensions, and to take it out a known measure of the 
lateral surface required. 
Eule. — Take the radius of the arc of the greater given base in 

the dividers, or beam compasses, and from any desired centre on the 

plate, from which the pattern is 
to be taken, as o, diagram, de- 
scribe an arc, as R v, to any suf- 
ficient length ; then with the ra- 
dius of the arc of the lesser 
given base in the dividers, or 
beam compasses, describe, from 
the same centre, a parallel arc, 
as r n, to any sufficient length ; 
then draw a right line from the 
central point to the outer arc, 
as o r R, diagram. Next apply 
the requisite measure to the arc 
of the base whose circumfer- 
ence is known, as from R to v, 

diagram ; and from the point v draw a right line to the central 

point, as v n o : the section it v n r will be the pattern proposed. 

Example. — The depth of a vessel is to be 10 inches; the 
diameter of one of its bases, 8 inches ; that of the other, 6 inch- 
es ; and a pattern containing one-half the lateral surface is re- 
quired. 




Now, by the foregoing formulae, R = vH 2 -\- (%Dy 
Dh 



and 



H = 



B — d" 



therefore 



R = J (d^) + (W- r > by the formula3 = 1} ; then 

1 /8 X 10\ 2 " ' 
I — ) -\- (|) 2 = 40.2 inches, the radius of the arc of the 

greater given base, and — '—- = 30.15 inches, the radius of the 

arc of the lesser given base. 

8 X 3.1416-1-2 — 12.5664 — 12^ inches, short; the required 
length of the arc of the greater base ; or 3.1416 X 6 -f- 2 = 9 T ^ 
inches, large, the required length of the arc of the lesser given 
base. 



PATTERN CUTTING. 



339 



Note. — As before stated, it is wholly immaterial which of the arcs ia 
known, for by defining one we define the other ; but generally the greater arc 
can be much more readily measured by mechanical means than the lesser, and 
it may be trimmed to facilitate the act, in which case the measuring-tape may 
be used. When the lesser arc is to be measured mechanically, the measure 
should be a strip of flexible plate, cut to the required length, and bent to the 
curve. 

If it is desired to convert the decimal part of an inch into eighths of an inch, 
multiply it by 8; if into sixteenths, multiply it by 16. Thus, the decimal 
.5664 X 8 = 4.5312 eighths as 9.0624 sixteenths. The decimal .5644, therefore, is 
practically equal to 9-16. 

Prob. 3. — To construct a Pattern for the Lateral Portion of a ves- 
sel in the form of a Frustum of a Cone, of given relative pro- 
portions or symmetry of outline, and given Capacity ; any two of 
its dimensions, and one of them a base, being given. 
The following table has been calculated for relative propor- 
tions as set down at the top of the columns, and for portions, P'n, or 
parts, of lateral surface, as set down in the left-hand column. 

R represents the radius of the arc of the given baseband D, 
the diameter of that base. Thus, if D be taken for the greater 
base, then it will represent the radius of the arc of that base, and 
the slant height of the cone from which the frustum is to be taken ; 
but if D be taken for the lesser base, then R will represent the 
radius of the arc of 'the lesser base, and the slant height of the 
cone that will be left after the frustum has been taken from it. 

H represents the perpendicular height of the cone having D for 
its base. 

c represents the chord of the required arc, or chord that will 
subtend the arc that must be on the pattern or portion set down 
in the left-hand column. 

S represents the cubic contents of the cone having D for its base. 



P'n 


R-D. 


R = HD. 


R=2D. 


R = 21D. 


R = 3B. 


R = iD. 


R = QD. 


RX—c 


RX=c 


RX=c 


RX=c 


RX=c 


RX=c 


RX=c 


1 


2 


1.8478 


1.4142 


1.1755 


1 


.7654 


.5176 


i 

± 


1.4142 


1.1113 


.7654 


.618 


.5176 


.39 


.2611 


1 


.7654 


.5176 


4158 


.3473 


.2611 


.1743 


.7654 


.5806 


.39 


.3129 


.2611 


.1961 


.1308 


4 
i 


.618 


.4669 


.3129 


.2506 


.2091 


.1569 


.1047 


| 


.5176 


.39 


.2611 


.2091 


.1743 


.1308 


.0872 


6 


#X=H 


DX=H 


DX=H 


DX=S 


DX=H 


DX=H 


DX=H 


q 


.866 


1.236 


1.9365 


2.4495 


2.958 


3.9686 


5.9791 


\tf t x=s 


D 3 X = S 


D 3 X = S 


n s x = s 


D3X=S 


n 3 x = s 


D 3 X = $ 


k .2267 


.3236 


.507 


.6413 


.7744 


1.039 


1.5653 



When R = bD, and pattern covering the whole lateral portion 
of the vessel is required, czz.GlSR; covering £, c = .3l29R; 
$,c = .209lR;%,c = A569R; \,c = .1257R; l, c = .1047/2; 
#=4.975D; S = 1.3024L^. 



340 



PATTERN CUTTING. 



General Explanation of the Table. — R — D; then 
72 X 2 is equal the chord of the arc that is equal to the whole cir- 
cumference of the base, and R X 1.4142 is equal the chord of the 
arc that is equal to half the circumference of the base, &c. 

R = 2D; then R X -7654 is equal the chord of the arc that is 
equal to half the circumference of the base, and Rx .5176 is 
equal the chord of the arc that is equal to one-third part of the 
circumference of the base, &c., 

The annexed diagram is drawn to radii 1, 1^, 2, 3, and 4 times 
the diameters of the ends, and exhibits symmetrical proportions" 
for vessels of this general class ; and, by help of the preceding 
table, a pattern containing the whole lateral surface of either 
special figure, or such portion thereof as set down in the left-hand 
column, may be readily and correctly obtained. 

When the greater Base of the Vessel is given. 
Rule. — Place the square suitably on the plate from which 

the pattern is to be taken, 
and scribe to its blades, as 
a m, m R. Make m R of 
sufficient length, and make 
m a equal to any desired 
aliquot part of the great- 
er given base ; then, with 
the square or straight-edge- 
in the position a R, and 
that measure 1, l£, 2, or 
more times the diameter 
of the greater base, ac- 
cording to the table and 
the special figure intend- 
ed, draw a line, or that 
measure, as a R. Next, 
if the perpendicular depth 
be given, space it off on 
the perpendicular, as m. h; 
and from h, with the square 
as at first dropped down 
to that point, draw a line, 
h c, parallel torn a, which 
will be the corresponding 
measure of the lesser base 
of the vessel. If the slant 
depth of the vessel, instead of the perpendicular, be given, space it 
off on the slant depth, asac; and from e, with the square as be- 




PATTERN CUTTING. 341 

fore, draw a line, c h, the corresponding measure of the lesser 
base, as before. If the lesser base of the vessel, instead of the 
perpendicular depth, or slant depth, be given, take the same rela- 
tive measure of that on the square that was taken of the greater 
base ; and with the square as at first dropped down on the perpen- 
dicular, until that measure is exactly included between the lines 
a R and m R, as h c, draw a line, h c, which will be the required 
measure of the lesser base. Next, with the radius a R, and R the 
centre, describe an arc, as a b, to any sufficient length ; and with 
the radius c R, and R the centre, describe a parallel arc, as c d, to 
any sufficient length. 

Suppose now, for example, that we have drawn the arcs to ra- 
dii once the diameters of the vessel ; that the diameter of the great- 
er base is 18 inches ; and that we would take out a pattern con- 
taining say \ part of the lateral surface required : then, on turn- 
ing to the table, we find in the column headed R—D, and 
against ^ in the column of portions, the constant, or co-efficient, 
0.7654 ; and we are told at the top of the column, that the 
radius multiplied by that co-efficient is equal the chord of the re- 
quired arc. Then 18 X -7654 = 13.779 inches, the chord of the 
arc that is equal to ^ part of the circumference of the greater 
base. Take, therefore, 13.779 inches in the dividers, and from the 
point a, with that measure, cut the arc, as at b, and from the new- 
found point, b, draw a right line to the point it, as b R ; then will 
the section a c d b be the pattern sought. This may be taken out, 
and marked with the values of R, d, and ^, with other distinguishing 
marks, as the depth, or capacity, or both, and kept for future use. 

Suppose again, for further illustration of the practical use of the- 
table, that we have drawn the arcs to radii 1£ the diameters of 
the bases ; that the diameter Of the greater base is 10 inches ; and 
that we would take out a pattern containing | the side of the' : 
vessel : then, on referring to the column headed R '== l^D, hi the 
table, we find in that column, against the portion % in the column 
of portions, the co-efficient 1.1113 ; and 10 X H X 1.1U3 =s= 14.82 
inches, the chord of the arc required. Take, therefore, 14.82 
inches in the dividers, or on the square, and from the point ^, with 
that measure, cut the arc as at b, and draw a line b R. The section 
a c db will be the pattern demanded. 

When the Lesser Base of tfre Vessel is given, and the Greater is 
unknown. 

Rule. — Place the square suitably on the plate from which the 
pattern is to be taken, as c h R, diagram, and draw to any suffi- 
cient length a line, h R; and from h, on the other arm of the 
29* 



342 PATTERN CUTTING. 

square, set off any aliquot part of the base, as h c, and draw that 
measure. Next, with the square, or straight edge, in position, as 
c R, and that measure 1, 1^, 2, or more times the diameter of the 
base, according to the table and the special figure intended, draw 
a line, as c R, and produce it sufficiently in the direction a. Next 
produce R h sufficiently in the direction m. Next, if the perpen- 
dicular depth is known, space it off on the perpendicular pro- 
duced, as h m, and, with the square raised up to the point m, draw 
a line, m a, parallel to h c, which will be the corresponding aliquot 
part of the greater base. If the slant depth is given instead of 
the depth, space it off on the slant depth produced, as c a, and, 
with the square raised up on the perpendicular till the shorter arm 
cuts the point a, draw a line, a m, which will be the required ali- 
quot part of the greater base. Next, with R a the radius, and R 
the centre, describe an arc, as a b, to any sufficient length ; and 
with R c the radius, and R the centre, describe an underlying arc, 
as c d, to any sufficient length. 

Suppose now, for example, that we have drawn the arcs to radii 
six times the diameters of the bases, that the diameter of the lesser 
base is 1 1 inches, and that we would take out a pattern containing 
one-third part of the side. On referring to the table we find, in 
the column headed R =. 6D, and against ^ in the column of portions, 
the constant .1743, and 11X6X0. 1743 = 1 1£ inches, the chord of 
the arc of the lesser base that is equal to one-third part of the circum- 
ference of that base. Take, therefore, 11^ inches on the square; 
and with that measure, as from c to d, cut the arc as at d; and from 
R, through the point d, draw a right line to the greater arc, as 
R db: then will the section a b d c be the pattern demanded. 

It may be proper to state that the table is also applicable in find- 
ing the sides or ends of a vessel in the form of a prismoid, or of a 
frustum of a pyramid, of any number of sides from three up, cor- 
responding to the denominators of the fractional portions tabulated. 

Prob. 4. — The special tabular figure, the Diameter of one end, and 
the Cubic Capacity of the vessel, being given, to find the Diameter 
of the other end, 

D, d = diameters of bases. 
a = cubic capacity of vessel. 
k = tabular constant for S. 

Example. — A vessel in the form of a frustum of a cone is to be 



PATTERN CUTTING. 



343 



constructed to radii 4 times the diameters of the bases : the diame- 
ter of the greater base is to be 7 inches, and the capacity exactly 
1 gallon (231 cubic inches). What must be the diameter of the 
other base ? 

Seeking k in the column headed R = 4Z>, in the table, and under 
I?X = S, we find it to be 1.039 ; then 



J 



7 3 X 1.039 — 231 t n „ n . . 

== 4.9416 inches. Ans. 

1.0o«7 



Having now both diameters of the vessel, proceed for a pattern 
containing the desired portion of the side, as directed in the fore- 
going. 

Example. — A measure of the exact capacity of 2 gallons is to 
be constructed to radii 3 times the diameters of the bases : the 
diameter of the lesser base is to be 5£ inches. What must be the 
diameter of the greater base ? 

k, by the table, = .7744 ; then 



J 



5.5 3 X .7744 + 462 «„„„„., 

^ „„,, ' = 9.1377 inches. Ans. 

.7744 



Example. — What will be the depth of the last-mentioned vessel? 

Seeking q in the column headed R — 3D, in the table, under 

DX=H,we find it to be 2.958 ; .-. # = 9.1377 X 2.958 = 27.0293 

inches, and h =H =- = 2.958 (D — d) = 10.76 inches. Ans, 

Example. — A pan in the form of a truncated cone is to be 
constructed to radii once the diameters of the bases : the diameter 
of the greater base is to be 15^ inches, and the capacity 8 wine 
quarts (462 cubic inches). What must be the diameter of the 
other base ? 

The constant for the solidity, in the column headed R = D, is 
.2267, and 

15.5 3 X .2267 — 462 



J : 



=11^ inches. Ans. 



Example. — What will be the slant depth, and what the perpen- 
dicular, of the aforementioned pan ? 

s = R — r, and h — lis — R. But R in this case = Z), and 
r=Rd-^-D = d; therefore s = 3.5976 inches. Ans. 

H, by the Table, = Z>X -866 = 13.423, and h = 13.423 X 3.5976 
-j- 15.5 =.866 (D — d) = 3.116 inches. Ans. 



344 PATTERN CtfTTING. 

By Mathematics. 
D,d = diameters of bases. 
R, r= radii of arcs of bases. 
H= perpendicular height of generating cone. 
~ h= perpendicular depth of vessel. 
S' = cubic contents of generating cone. 
a = cubic capacity of vessel. 



D = 

d = 



J12S_ | Sd 3 _ l/ 3 12da \ 



.._ 125 12Da . 12a 



nD* ir(D*-d 3 y ~ ntDd+W+d?) 



_ nim D*a izJ^h 



ftfef^ 



12 D 3 — d 3 12(Z> — d) 

_ S(D* — d*) = nH(D\— d s ) 7th(Dd -j- D 2 -f d*) 
a ~ D 3 .".,; 12D ~ 12 

For other forms of expression, and other applicable equations, 
see p. 331. 

Prob. 5. — To construct a Pattern for the Lateral Portion of a ves- 
.... sel in the form of a Frustum of a Cone, of given Tabular Outline, 
and given Dimensions, without plotting the dimensions. 

'■■■ s = slant depth of vessel. 

g = tabular ratio of R to D = 1, 1|, 2, 3, &c, top line. 
& q = tabular ratio of H to D, third line from bottom. 

k = tabular constant for S, bottom line. 

y z=z tabular constant for chord of arc. 

Other symbols as in the foregoing. 

When both Bases of the Vessel are given, 
R = Dg, and r = dg; h = q(D — d), s = R — r. 
When the Greater Base and the Depth are given, 

* q R g q 



PATTERN CUTTING. 345 

When the Greater Base and the Slant Depth are given, 
R == Dg, and r = R — s. 

When the Lesser Base and the Depth are given, 

R= r(dq±K) D== ™ = ? = <ll±A. 

dq y r g q 

When the Lesser Base and the Slant Depth are given, 
R = r -J- s ; r = dg. 

Example. — Given the tabular outline R = 2 D, the diameter 
of the lesser base d=5, and the depth h = 8, to find the diame- 
ter of the greater base D, the radii of the ares R, r, and the 
chord of the greater arc c, that will subtend ^ the circumference 
of the greater base. 

By the Table, when g = 2, and p = £ ; that is, when R = 2 Z>, 
and a pattern containing one-half the lateral surface is required, 
g = 1.9365, and y=z. 7654; 

18 26 
Dz=-^~ =9.13; and c = 18.26 X .7654 = 13.976. Am. 

Example. — Given D = 10, s = 4, g = 3, p = £, to find R, r, 
and c. 

By the Table, qz= 2.958, and?/ = .3473; theni2=10X 3 = 30; 
r = 30 — 4= 26; and c = 30 X .3473 = 10.419. Arts. 

When the tabular outline R = Dg, the assumed diameter, and 
the cubic capacity only, are given, the other diameter must be 
found, and by one of the following formulas, viz. : — 

D== \ d * kJ r a 

k 



d = 



■j('+s> 



Rule. — Take R in the dividers, or beam compasses, and from 
any suitable centre, o, describe an arc, R v, to any sufficient length. 
Next, with r the radius, and the same centre, describe an arc par- 
allel to the former, r n, to any sufficient length ; then draw a right 
line from the central point to the outer arc, and cutting the arcs, 
as at r and R. 




346 PATTERN CUTTING. 

Suppose, now, for example, that the arcs are drawn to radii once 
and one-third the diameters of the bases ; that the radius of the 
greater are is 16 inches ; and that you wish to take out a pattern 
containing one-fifth part of the side of the vessel. Referring to the 

table, column R = l^D, and 
against l in the column of por- 
tions, you find .4669; and 
16 X -4669 == 7.47 inches, the 
chord required. Take therefore 
7.47 inches in the dividers or 
on the square, and from the 
point R, with that measure, cut 
the greater arc as at v, and from 
the new-found point, v, draw a 
right line to the central point 
o, and cutting the lesser arc, 
as at n ; then will R v n r be 
the pattern sought. 

On the contrary, suppose you have drawn the arcs to radii four 
times the diameters of the bases, that the diameter of the lesser base 
is 5 inches, and that a pattern containing one-third part t)f the lat- 
eral portion of the vessel is required : then, by the table, column 
R = 4D ; and against \ in the column of portions, the constant, .261 1 
appears ; and 5 X 4 X .2611 = 5.222 inches, the chord of the arc 
of the lesser base that is equal to one-third part of the circumfer- 
ence of that base. Take, therefore, 5.222 inches on the square, 
and from the point r, with that measure, cut the lesser arc, as at n ; 
and from the central point o, through the point n, draw a rigjit line 
to the greater arc, and cutting that arc, as at v : then will the sec- 
tion R rn v be a pattern for the vessel, and contain one-third part 
of the side. 

Prob. 6. — The Capacity in gallons of a Vessel in the form of a 
Frustum of a Cone being given, and any two of its dimensions, to 
find the other Dimension. 

It has been shown that — X- — — — - = solidity, or capa- 

l £ 
city, of a frustum of a cone, the solidity and the 'dimensions being 
in the same terms of measurement ; 



~~^\7rh a) 2* 



But since a gallon contains 231 cubic inches, and 
it 



4 X 231 



= .0034, 



PATTERN CUTTING. 347 

it followa that •°°«**( J *+ g + <P ) == .OOUi>(J tf + J i + J) 

o 

= solidity, or capacity in gallons ; the dimensions being in inches. 
Therefore, putting C to represent the capacity of the vessel in 
gallons, 

.oo34(2>rf + z) 2 + rf 2 ) ; ^Jvoosa 4 / 2 ; 

^J\.0034A 4 / 2* 

Example. — A measure in the form of a frustum of a cone is to 
be constructed to the exact capacity of 3 gallons ; the diameter of 
one of its bases is to be 11 inches, and the perpendicular depth 1 2 
inches. What must be the diameter of the other base ? 



Jo 



8X> II- XH 11 u,^. Am . 



k l2 X -0034 4/2 

Example. — A vessel in the form of a frustum of a cone is to be 
constructed to the capacity of 3 gallons ; the diameters are to be 11 
inches, and 8| inches. What must be the depth ? 

.0034(11 X8 3 7^+ll' + 8.75») =!M)liDCheS - Am - 

Example. — A measure in the form of a truncated cone is to be 
constructed to the capacity of \ gallon ; the diameter of one end is 
to be 5^ inches, and that of the other 4 inches. What must be the 
depth ? 

.0034(22 +1o1« + 16) = M64meh-. %** 

Example. — A measure is to be constructed to the capacity of 
one wine quart ; the diameter of one of the bases is to be 4 inches, 
and the depth 5^ inches. What must be the diameter of the other 
base ? 



Jc 



L )-r=" 



3Xi = -'5 3X4* 
0034 X 5.0 —) - 5 =«*»**»■ *»• 



Example. — A measure of the capacity of one wine pint is to be 
constructed ; the depth is to be 4^ inches, and one of the diameters 
is to be 3^ inches. What must be the other diameter ? 



JC 



'-M^-'-^T 3 ) -1-^=2.1644 inches. Ans. 



348 RATTEEN CUTTING. 

Example. — A measure is to hold \ wine pint, and each base is 
to be 2f inches in diameter. What must be the depth ? 

3X T V .0625 .0625 '„„*„•' , 

.0034 X 3^ = ^034^ = .0034 X 2 .375^ 3 ' 259 mcheS ' AnS ' 



Prob. 7. — To construct Patterns for flaring oval Vessels of differ- 
ent Eccentricities and given Dimensions. 

The solids here contemplated, and for which coverings are to be 
constructed, closely resemble the frustums of elliptic cones ; but 
they are not identical with that figure, since an oval is made up of 
circle arcs, while no part of an ellipse whatever is strictly the arc 
of a circle. The leading principle, however, for this class of pat- 
terns, so far as regards their sides, is expressed by the act of an 
elliptic cone rotating on its side, on a plane, from the line where 
the plane of the transverse axis of its base is at right angles to the 
plane on which it rotates, to the extent of one revolution of the cone. 

D represents the transverse diameter, and d the conjugate, of the 
greater base ; D f the transverse, and d' the conjugate, of the lesser 

Dh 
base ; h the perpendicular depth of the vessel : Hz= jy — ^ the 



perpendicular height of generating cone ; M= y//? 2 -\~ (J-D) 2 , the 

maximum slant height of generating cone ; S= ^H 2 -\- Oj^) 2 , 
the minimum slant height of generating cone. 

No. I.— The Lateral Portion. D : d\ \ D' ; d f . 

tf=F; ta = P;fa = R;fc=N; fv=S:fA=M; dia- 
gram. 

Rule. — Place the square suitably on the plate from which the 
attern is to be taken, and scribe to its edges at, t R, making both 
ines of sufficient length ; and produce t R sufficiently in the°direc- 

= 0.345092A and draw fa. Make t i = \h*+ ( D — D ')\ and 

^ 12 

i>2 on one blade 

of the square, and the limit of that measure in the point/, the 
other blade cutting the point a, as/c a, draw/c, c a, and produce 



I 



PATTERN CUTTING. 



349 



fc in the direction A. Make t R =F 



Make 



MN 



and draw a R. 



% r = 




D'(t R) 
— XT"' 
and draw m r, 
which will be par- 
allel to a R. Then 
with a R in the 
dividers, and R 
the centre, de- 
scribe a v; and 
with m r in the 
dividers, and r the 
centre, describe m 
k; also with a y 
the radius, and y K 
the centre, de- 
scribe a A ; and 
with m z the ra- 
dius, and z the 
centre, describe wi 
C, Thus A avk 
m C will be the 
unit measure of 
the pattern, and 
will contain one- 
fourth part of the 
side. 



Note. — It will be perceived that the section on the right of the per- 
pendicular v R, diagram, is an exact duplicate of that on the left, only reversed ; 
thus A v B D h C is a pattern for the vessel, and contains one-half its lateral 
surface. The last-mentioned pattern may be obtained by draft, by repeating 
the lines and arcs on the left; or it may be obtained by scribing to the unit 
measure in the two positions. In making up, the side should consist of a 
single piece when practicable, and the lock or seam should be under the 
handle. 

To construct the Bases for No. 1. 

Rule. — Lay off A B, diagram, equal in length to the trans- 
verse diameter of the base, and divide it into three equal parts, e 
and r; also bisect it, and through the point of bisection, o, draw an 
.■■■■■, 30 



350 



PATTERN CUTTING. 



indefinite perpendicular, as C D. 



Make e p, eg, each equal to 
A e or B r ; and from 
the points p and g, 
through the points e 
and r, draw right 
lines, as g e s, g r t, 
p e m, p r n. Then 
with A e or B r the 
radius, and e and r 
the centres, describe 
m A s, t B n ; and 
with g s or p n the 
radius, and g and p 
the centres, describe 
s t and m n. 



By Mathematics. 

4 — - yd o 

= 0.7559831.0; 
op = og=z sy = -^- 2 = 0.28S675D;pC=zgD=i ( 2 ~^ D 

= 0.0893164D; 

3 




ps — ep = A e = oy = - 
A y — e y = or — — ; s g t = 60° ; m e s=: 120°. 



gsz=A r~g C = —- ; 



iVo. 2.— The Lateral Portion. D\d\\ U \d f . 

D represents the transverse, and d the conjugate, diameters 
of the greater base ; D' the transverse, and d f the conjugate diam- 
eter of the lesser base ; h the perpendicular depth of the vessel. 
TT __ Dh, 
" — r>_ /V tne perpendicular height of the generating cone ; 

S === \fH' 2 -\- {^d ) 2 , the minimum slant height of generating cone ; 

M= \/H 2 -\- (%Dy, the maximum slant height of generating cone. 

tf=F; ta = P; fa — R; fc — N; fv=S; 
fA = M; diagram. 

Rule. — Place the square suitably on the plate from which the 
pattern is to be cut (a t R, diagram), and draw lines a t, t R, each 



PATTERN CUTTING. 



351 



of sufficient length ; and produce R t sufficiently m the direction v 



Make*/= H 2 -j-~ andt 



|4^ + d 2 



= 0A21637D, 



and draw f a. Make £ i = I A 2 J_ 



allel to t a. Next, with cf 



>J 20 



(D—pry 



j and draw i 



par 



on one edge of the square, 



the limit of that 
measure in the 
point f, the edge 
of the other blade 
cutting the point 
a, as f c a, draw 
fc, c a, and pro- 
duce f c in the di- 
rection A. Make 

and draw a R. 
Make 

,_VJtK} 

"- D~' 
and draw wi r, 
which will be par- 
allel to a R : then 
with a R the ra- 
dius, and R the 
centre, describe a 
v ; and with r m 
the radius and 
r the centre, de- 
scribe m k: also 
with a y the radi- 
us, and y the cen- 
tre, describe a A; 
and with m z the 
radius, and z the 
centre, describe 
wiC. Then will 
A a v km C be the unit measure of the lateral portion of the ves- 




352 



PATTERN CUTTING. 



Rule. 



sel, and contain one-fourth part of that portion. See Note at the 
close of the directions for constructing the lateral portion of No. 1. 

To construct the Bases for No. 2. 
Make A B equal to the transverse diameter of the base, 

and divide it into four 
equal parts, in e, o, 
and r ; and through o 
draw a perpendicular, 
C D, of sufficient 
length. Make o C, 
o D, each equal to one- 
third of A B, and from 
C and Z>, through e 
and r, draw C e m, 
Cr n, D e s, D r t. 
Then with D C the 
radius, and D and C 
the centres, describe 
s C t, and m D n ; also 

with A e or B r the radius, and e and r the centres, describe 

m A s and t B n. 




AB^D = ^;d = ~;oC 



By Mathematics. 
D 

3 



oD= — ; o p = s y — 



2D 



2D 



D 
5 

SD 



10'^ 15 



s Z)*=73°.74; trn = 106°A. 
No. 3 — The Lateral Portion. D'.D'y.d'.d'. 
Symbols, same as for No. 1 and No. 2. 

Rule. — Draw at,t R, at right angles to each other, and of suf- 
ficient length ; and produce R t sufficiently in the direction v. Make 



*/ 



3D 2 
H* + — ,' and t a = ^.15Q72UD* = 0.3882285Z), and 



draw /a Make £ i = ft 2 -L. -i ^ and draw i m parallel to 

I w 

a t Next, with c f — \R 2 on one blade of the square, and 

the limit of that measure in the point /, the other blade cutting 



PATTERN CUTTING. 



353 



the point a, as fc a, draw fc, c a;, and produce / c sufficiently 

MN 
in the direction A. Make t R = F-\ p—> and draw a R. Make 

ir= — t)~~ ' an( ^ draw m r, which will be parallel to a R. Then 

with a R the radius, and it the centre, describe the arc a v; 
and with r m the radius, and r the centre, describe the arc m k ; 
' also with a y the radius, and y the centre, describe the arc a A ; 
and with m z the radius, and z the centre, describe the arc 
m C : then will the section A a v k m C be the unit measure of 
the pattern, and contain one-fourth part of the side. 



To construct the Bases for No. 3. 

Eule. — Make A B equal to the transverse diameter of the 
base, and divide it into four equal parts, in c, o, r ; then bisect it 
with a perpendicular of suf- 
ficient length. Make e i, 
e g, each equal to half A B, 
and from the points i and g, 
through the points e and r, 
draw i e m, i r n, g e s, 
g r t, all of sufficient length. 
Next, with A e or B r the 
radius, and e and r the cen- 
tres, describe the arcs m A s, 
t B m; and with g s or g t 
the radius, and g and i the 
centres, describe the arcs 
s C t, m D n. 




AB = D = 



2d 



By Mathematics. 

— 1.578Ud; d = ( 3 —^ 3 ) D — .6339746D; 

3D 



ps=oy = —- 1 



op 



_Z>y/3 
~~ 8 



■21650635Z) 



; pC=S W*=*)£ = . 



8/3 



100481D; 



D 



Di/3 3D 

oiz=zog=z—%- ; g S • =zAr=z—^; sgt = 6Q°; trn=zl20°. 



354 PATTERN CUTTING. 

The Lateral Portion by Mathematics, be the Eccentricity what it may, 
p s = s, sine of the arc s C, base ; 

op = g, sine of the arc s A, base (see diagram of base). 
D, D', d, d', h, H, as in the foregoing, D \ D' y.d'.d'. 

fv=S = y/lH* + (**)■] ; fA == M= ^ + (*!>)■] ; 
ca=p = ^(R*-N*); cy=f=FN+(tR)=^(r*- P *); 

«3/=^ = v / (/ 2 +i> 2 ); ti=y/(v+ 9 * {D & D,) y 

mr= \/[(i rf + (i m) 2 ] = -1?J9 ; im = yTO r Y ~ (* r ) 2 ] 

- vTO/) 2 - (if 2 )! = ^; »> = vTOO 2 - (*•»»)«] =^^ ; 

MN 
tR = F + -jf-= ) /l(aRy-I*}i aR = ft(t-Ry + I"]i 

mz = r' = D'r+D; xm=p' = D r p-^-D; xz=f , = D , f~D; 
ik = D'(S — F)-±-D; fi = F—(ti); Cx = D'(M—N)+D; 
cAz=. M — N ; tv = S — F; also, I or A a v, diagram, 

JTtDd (D-d)* * 
learly ; and 4Z -f- 7r = diameter of generating circle of ellipse. 



0.8825 



Of Cylindrical Elbows. 

The solid to be taken into view in reference to the construction 
of the arms of a hollow cylindrical elbow is a cylinder of the di- 
ameter proposed for the arms, having one of its bases oblique to 
the sides to the extent of half the angle proposed for the elbow; 
and the best mode of laying off the arms, generally speaking, is 
expressed by the act of this cylinder rotating on its side on a plane, 
from the line where the plane of the transverse axis of its oblique 
base is at right angles to the plane on which it rotates, to the ex- 
tent of one revolution of the cylinder. 

* This formula for one-fourth of the perimeter of an ellipse affords almost 
strict accuracy when the conjugate diameter is not greater than two-thirds, nor 
less than one-third, of the transverse. It furnishes it too short by 1-H00 
when the conjugate is equal to three-fourths of the transverse, and too long 1 
by 1-1100 when the conjugate is equal to one-fourth of the transverse. See 
Conic Sections, Ellipse. 



PATTERN CUTTING. 



355 



It is apparent, then, that the true face, or outline, of the joint 
of a cylindrical elbow, when the cylinder in continuance is rolled 
into place, is an ellipse, whatever be the angle of the elbow. And 
it may be shown that if a hollow cylinder having an oblique base 
be bisected by the plane of either axis of that base, and the semi- 
cylinders opened to plane surfaces, the portion of the curve that 
will be on each will be a hyperbola, or two equal and similar semi- 
hyperbolas united, and forming a cima, or facing in opposite direc- 
tions, according as the plane of the conjugate axis, or that of the 
transverse, be made the cutting plane alluded to ; and it is clear 
that the curve that will be on one piece will be equal and similar 
to that on the other in all particulars. Thus the curve of the 
joint of a cylindrical elbow, when laid off on a plane, is made up 
of four equal and similar semi-hyperbolas for either arm of the 
elbow ; the method of locking the joint not now being taken into 
account. 

From the foregoing analysis, the following rules and directions 
for practice have been derived ; and it is believed they will be found 
not only as correct and simple, but as ready of execution, as any 
that have been or can well be devised for the purposes proposed. 
Prob. 8. — For a Right-angled Elbow, 
d = diameter of pipe, 
d X .7854 ±= \ circumference of pipe. 

Rule. — Construct on any plate suitable for the purpose, and 
for taking out, a rectangle, A C B D, in length, A D, equal to 
one-fourth part of the cir- 
cumference of the pipe, and 
in breadth, A C, equal to 
one-half the diameter of the 
pipe. Make A e equal to 
dX 0.5708, and divide the 
space into any number of 
equal parts. Divide CB 
and B D each into the same 
number of equal parts that 
you divide A e into. Con- 
nect the points of division in 
A e and C B directly, by right lines, as 1 1, 2 2, 3 3, &c. Then, 
from the points of division in B D, draw right lines tending direct- 
ly to the point A, as 14, 2 A, 3 A, &c. The intersections of the 
lines bearing like numbers will be so many points in the locus 
(place or line) of the required curve ; and the more numerous 
these, the more completely, of course, will the curve be defined. 
The practice may be called locating the curve by intersecting lines, 




356 



PATTERN CUTTING. 



and is equivalent to locating it by ordinates. Trace the curve, and 
the unit measure, A B C, of the true arm of the contemplated el- 
bow will be constructed, and may be taken out for use. 

Prob. 9. — To apply the Unit Measure to the Construction of a 
Full Pattern for the Arms of the Elbow. 

Rule. — Draw a guide-line, C a C, suitably on the plate from 
which the pattern is to be taken, and of sufficient length. Then, 
with the measure in position 1, scribe to 
its sides C A, A B ; with it in position 2, 
scribe to its sides B A, A a; with it in 
position 3, scribe to its curve A B ; and 
with it in position 4, scribe to its sides 
B A , A C. This practice, care being taken 
during the proceedings to keep the base 
of the measure directly in the guide-line 
C a C, and that the extremities of the 
curve are made to unite in the same 
points, will construct the curve AAA, 
the curve proposed. 

Now, as may be readily inferred, were 
the workman to cut out both arms of the 
elbow by this curve, and then to roll them 
properly into cylinders, they would unite 
by their oblique bases uniformly through- 
out ; and would form a right angle, or 
stand to each other at an angle of 90 de- 
grees. Moreover, were he to turn a burr 
on the oblique base of one of them, and 
a ledge and lip to match on that of the 
other, both parallel in their practical bear- 
ings to the plane that is common to both 
the said bases, the same state of things 
would still be maintained ; and the arms would lock with a close 
uniform flange bisecting the angle. 

We have thus far spoken of the true or geometrical cylindrical 
elbow ; but we come now to speak of the arms as they ordinarily 
come from the edging machine, which, short of the strictest han- 
dling with reference thereto, can scarcely be made to turn the wards 
as above proposed, and, moreover, the flexibility of the plate is 
usually insufficient to admit of it. 

The work, however, as it ordinarily comes from the edging- 
machine, lacks uniformity and definiteness in nearly all particulars. 
The wards, beside being out of parallelism to the plane that is 
common to the oblique bases, are more or less irregular, and out of 




PATTERN CUTTING. 357 

harmony with each other. Scarce any two workmen turn them 
alike ; and scarce any workman, it may almost or quite be said, 
ever turns them twice alike. The condition of the machine at one 
time compared with another ; the thickness of the plate employed ; 
the customary manner of handling the work in the machine ; and 
the fact that one of the arms has usually to be made less in circum- 
ference at its extremity than at the joint, that it may enter the 
flush end of a straight pipe of the same nominal diameter, which 
tends to displace the curve on that arm, — these circumstances, I say, 
render the irregularity and indefiniteness spoken of unavoidable, 
in a greater or less degree. 

There is one point, however, which we can fix upon, viz., — the 
wards come from the edging-machine invariably less oblique in 
their practical bearings to the sides of the 
cylinders than comports with the angle 
proposed for the elbow, unless ttiat angle 
is considerably greater than 90 degrees. 
, It is the practice with many workmen 
to retain the true curve for the outside arm 
of the elbow, and after closing the straight 
lock to turn its ward unhesitatingly upon 
it ; and then, after closing the straight lock 
of the inside arm, and before turning the 
burr, to cut away or trim its upper limb, 
or crown, until it will properly enter and 
lock. But this practice does not tend to 
correct the angle ; and unless the wards 
are turned well down in the throat, and 
rather narrow at the crown, the angle will 
be greater than demanded. 

The best general rule that I have been 
able to put to a practical test for construct- 
ing a general pattern for both arms of the 
elbow, of from 4 to 8 inches in diameter, 
is the following : — 

Rule. — Construct the curve AAA, 
with its lateral and central guides, as al- 
ready directed; then with the measure 
successively in positions 1, 2, 3, 4, but dropped down equally in 
each position, according to the desired width of the pattern, con- 
struct the curve a a a; taking care the while to keep the perpen- 
dicular of the measure directly in the guide-lines, C A, produced ; 
and that the extremities of the measure unite in the same points. 
This will duplicate the true curve, and place the curves parallel in 
position. At this stage, it will be well to make the central line 




358 PATTERN CUTTING. 

A a distinct and permanent, and to drop the permanent perpen- 
diculars B b, B b. Next, make A m, A n, each equal to one-fourth 
of an inch, and make a i, a e, each equal to one-eighth of an inch. 
Then, with the measure nearly in position 1, its angle B in the 
point B, and its curve tending to the point m, scribe to its curve, as 
B m ; and with its angle B in the point &, and its curve cutting 
the point i, scribe to its curve i b ; also, with the measure nearly in 
position 4, its angle B in the point B, and its curve tending to 
the point n, scribe to its curve, as B n ; and, with its angle B in 
the point b, and its curve cutting the point e, scribe to its curve e b : 
then m A n e ai will be the pattern contemplated. 

Prob. 10. — To apply the Pattern to the Construction of the Arms 
of the Elbow, Prob. 9. 
Rule. — Place the pattern suitably on the plate from which the 
arms are to be taken, and scribe to its curve m An; then raise it 
up on the plate till the points b b are in the points B B, and scribe 
to the curves i b,eb: then the curve m A n, with the continuation 
of the plate, will be the outside arm of the elbow, and the curve 
i A e, with the continuation of the plate, will be the inside arm. 

Prob. 11. — To find A C of the Unit Measure for any Angle of 
Elbow whatever. 
Rule. — Draw three sides of a parallelogram, as A C, A D, 
D B, diagram. Make A D equal to half the diameter of the arms, 
and make A C, D B, each of sufficient length ; then, with the bevel 
square set to half the angle proposed for the elbow, as C A B, and 
one of its arms directly in A C, scribe to its arm A B : then will 
D B be equal to A C required. 

By Mathematics. 

For the Unit Measure of an Elbow of any given Angle. 

dz=z diameter of arms = conjugate axis of hyperbola. 
A D = a zr:7rrf-f-4 = .7854c? — % circumference of arms. 

V= angle of elbow in degrees. 
A e = 2a — d=±ind — d=.5708d. 

AC-b=* .=*^-# 

tan h V sin £ V 

db 



t = transverse axis of hyperbola =~rQd-{- y/a 2 -}- ^c? a ) 
d 



x ■=. abscissa from vertex = -yV(|c? 2 -\-Jf) — £t 



d 
y ss ordinate = — tf(tx-\- ar 2 ). 



PATTERN CUTTING. 



359 



Prob. 12. — To construct a Right-angled Elliptical Elbow. 

Rule. — Construct a rectangle, A D B C, in length, A D, equal 
to one-fourth part of the circumference of the ellipse, or elliptic 
collar, and in breadth, A C, 
equal to half the conjugate 
diameter of the ellipse. 
Make A e = 0.18169 multi- 
plied by the circumference 
of the ellipse, and in all oth- 
er respects construct the 
unit measure by rule, 
Prob. 8. 

Next construct the full 
curve, A A A, by rule, 
Prob. 9, and take out both 
arms by that curve ; then lock the arms by their straight locks, 
turn the proper ward on the outside one, and trim the inside one, 
if necessary, to match. 

Prob. 13. — To find A C of the Unit Measure of an Elliptic 
Elbow of any given Angle. 

Rule. — Make A D equal to half the conjugate diameter of the 
ellipse, and in all other respects proceed by rule, Prob. 11. 




By Mathematics. 
For the Unit Measure of an Elliptic Elbow of any given Angle. 

C = circumference of ellipse. 
V=. angle of elbow in degrees. 
c = conjugate diameter of ellipse. 
d= C-^-n= conjugate axis of hyperbola. 
A e = ±C(TT—2)-±-iT=±nd — d. 
A Dz=za = \ circumference of ellipse. 
£c %c X cos £ V 



A C=b = 
db 



tan£F 



sin£F 



t = — j [&d-\- ^( a> + ?^ 2 )] = transverse axis of hyperbola 



x — & V(%d 2 -f- y 2 ) — 2* = abscissa from vertex. 



y = — tf(tx-\- z 2 ) = ordinate. 



360 



PATTERN CUTTING. 



Pkob. 14. — To construct ilie Quadrant of a given Circle by inter- 
secting lines. 

Rule. — Construct a square, A B C D, making each side equal 
to the radius of the proposed circle, and make A e equal to the di- 
ameter of the circle ; thence by rule, Prob. 8. 



Prob. 15. — To construct the Quadrant of a given Ellipse by in- 
tersecting lines. 

Rule. — Construct 
arectangle,y4 B CD,' 
making A B equal to 
half the transverse 
diameter, and A D 
equal to half the con- 
jugate. Make A e 
equal to the conju- 
gate; thence by rule, 
Prob. 8. 




Prob. 16. — To apply the Quadrant of a Circle, or Quadrant of an 

Ellipse, to the Construction of the Circle or Ellipse. . , -\ 

Rule. — Draw a guide-line of sufficient length ; then, with A B 
of the quadrant in that line, in the four requisite positions, scribe 
to the curve and to B C, as required. 

Prob. 17. — To construct the Quadrant of a Cycloidal Ellipse by 
intersecting lines. 

Rule. — Construct a rectangle, A B C D, making the trans- 
verse diameter to the conjugate as n to 2 ; in all other respects 
proceed by rule, Prob. 15. 

Prob. 18. — To describe an Ellipse of given diameters, by means 
of two Posts, a Pencil, and a String. 

Let A B be the transverse diameter, and C D the conjugate. 
Rule. — Lay down the given diameters at right angles to each 



/ 



PATTERN CUTTING. 361 

other, and bisecting each other, as in o. Next, with half A B in 
the dividers, and one 
foot in C or Z>, cut 
A B in y and z. Next, 
strike in a pin at y and 
another at z. Next, 
pass a string around 
the pins, and tie it at 
such a length that, the 
loop may be extended 
to C or D. Next, in- 
troduce a pencil, and, 
bearing upon the 
string, carry it around 
the centre, to the con- 
struction of the full 
circumference. 

Note. — If we let I represent the circumference of an ellipse, D the trans- 
verse diameter, and d the conjugate, then 2 = 3.53 \ \~\ ' ) ' 

very nearly. And I may add, that the expression affords almost strict accuracy 
when d is not greater than two-thirds nor less than one-third of D. It gives I 
too long by 1-274 when d=z±D, and too short by 1-351 when d=|Z). (See 
Conic sections, Ellipse.) 

Prob. 19. — To construct a Semi-parabola by intersecting lines. 

Rule. — Construct a rectangle, A B C Z>, making A B equal 
to the altitude of the parabola, and B C equal to half the base ; 
make A e — B C—A D ; thence by rule, Prob. 8. 

Prob. 20. — To construct a Right-angled Circular-Elbow of 3, 4, 
5, 6, 7, or 8 pieces, of any given Diameter, arid any given Radius 
of Curve. 

d = diameter of pipe. 

r = radius of throat (usually = \d, %d, -|cZ, or d). 
ird~ 4 = .7854^ = i circumference oi pipe. 
V= angle at centre in degrees. 



C = 



The Unit Measure, 
d X tan V 



2 

A D=C B = .7S5id. 
A e = .6708d. 

31 



362 



PATTERN CUTTING. 



No. of 


Angle at 


A a = 


Ab — 


ba = 


A C= 


pieces. 


Centre. 


(d-tr)X 


d X tar. V 


rX 


d X i tan V 


3 


22° 30' 


.41421 


.41421 


.41421 


.207105 


4 


15° 


.26795 


.26795 


.26795 


.133975 


5 


11° 15' 


.19891 


.19891 


.19891 


.099455 


6 


9° 


.15838 


.15838 


.15838 


.079190 


7 


7 a 30' 


.13165 


.13165 


.13165 


.065825 


8 


6° 25' 43" 


.11266 


.11266 


.11266 


.056331 




Rule. — Construct a rectangle, A C B D, in length, A D, equal 

to one-fourth part of the circumference of the pipe, and in width, 

A C, equal to \d X tan V. 
Make Ae = dX 0.5708, and 
in all other respects proceed 
by rule Prob. 8. 

Suppose the unit measure 
of a right-angled circular el- 
bow of 3 pieces is required, 
and that the diameter of the 
pipe is to be 7 inches ; then, 
On turning to the table, col- 
umn headed 

A C=dX £tan F, 

and opposite 3 in the column containing the number of pieces, we 

find the co-efficient, or half the tangent of 22° 30' to be 0.207105 ; 

therefore A C of the unit measure = 7 X 0.207105. = 1.449735 in.; 

A D= 0.7854 X 7 = 5.4978 in.; and A e = 0.5708 X 7 = 3.9936 

inches. 

Suppose the unit measure of an elbow of 5 pieces is required, 

and that the diameter of the pipe is to be 8 inches, then 

A D = .7854 X 8 = 6.2832 inches ; 
A C = S X .099455 = .79564 inch; 
A e = 8 X -5708 — 4.5664 inches. 

On the contrary, suppose the unit measure of an elbow of 4 
pieces is required, and that the diameter of the pipe is to be 
6 inches, then 

A D = .7854 X 6 = 4.7124 inches ; 

A C= .133975 X 6 = .80385 inch ; 

A e = .5708 X 6 = 3.4248 inches. 



PATTERN CUTTING. 



363 



Prob. 21. — To apply the Unit Measure to the Construction of the 
several Segments of a Circular Elbow. 

Rule. — Draw a guide-line C C, equal in length to the circum- 
ference of the pipe, and construct the curve, b A &', by scribing. to 
the measure in its several positions, 1, 2, 3, 4, as by rule, Prob. 9. 
This will make A C= C b=^d X tan V, and will make 
Ab = 2(C b) = dX tan V. Make b a, b' a', each equal to 
r X tan V, and draw the line a a', which will make A a, A' a', 
each equal to (</-}- r ) X tan V. Thus b A V a ! a will be the unit 
segment of the elbow ; and, with the requisite continuation of the 
plate, will form one of the two equal and similar outside pieces of 
the elbow. The inside segments are equal one with another, and 
each is equal to two of the outside segments, in all cases. 

Suppose a right-angled circular elbow of 3 pieces is required ; 
that the diameter of the pipe is to be 7 inches, and that the throw 
or radius of the throat, r, is to be 6 inches ; then 

C C f or aa r of the segment = 3.141 6 X 7 = 21.9912 inches. 
A C=Cb = dX itan 22 9 30' = 7 X .207105 = 1.449735 in. 
b a = r X tan 22° 30' = 6 X .41421 = 2.48526 in. 
ia=(dfr) tan 22° 30' = (7 -J- 6) X -41421 = 5.38473 in. 

The middle segment, therefore, of a 3-piece cir- 
cular elbow will take the form No. 2, or No. 3, 
following diagram (No. 2, with reference to econ- 
omy of stock, when the outside segments are 
made to lock at the throat). Thus the maximum 
width of the middle piece of a circular elbow con- 
sisting of 3 pieces = 2 (A a) = 2(d -f r) tan 22° 
30' ; and the minimum width of the same piece 
= 2 (b a) = 2r X tan 22° 30'. 

Suppose a 5-piece elbow is required; that the 
diameter of the pipe is to be 6 inches, and that 
the throw, or radius of the throat, is to be equal 
to § the diameter, or 4 inches ; then 

C C f = 3.1416 X 6 =18.8496 inches; 
A C=Cb = .099455 X 6 = .596 73 inch; 
b a = .19891 X 4 = .79564 inch ; 
ia=(6-f4)X .19891 = 1.9891 inches. 

On the contrary, suppose that a 6-piece elbow 
is required ; that the diameter of the pipe is to be 
8 inches, and that the radius of the throat is to 
be equal to %d, or 4 inches ; then 

CO — 3.1416 X 8 = 25.1328 inches ; and, by 
the table, 




364 



PATTERN CUTTING. 



A C—Cb= .07919 X 8 = .63352 inch ; 
b a — .15838 X4 = .63352 inch ; 
A a — .15838 X (8 -f- 4) = 1.90056 inches. 
From the foregoing it may be perceived that to cut stock with 

reference to economy, 
we must make use of 
the three annexed gen- 
eral forms for the seg- 
ments of a circular 
elbow whenever the 
elbow is to consist of 
more than three pieces ; 
whereby one or more 
of the segments will 
have its straight lock at 
the crown r and the oth- 
er, or others, at the 
throat ; and, thus, an 
elbow consisting of an 
odd number of pieces 
will make up without 
waste. 

The two outside seg- 
ments of a circular 
elbow are equal and 
similar, one to the other, 
in all eases ; and, if 
they are to have their 
straight locks at the 
throat, are but exact 
copies of the unit segment above described - y thus, the segment 
b A b' a f a, preceding diagram, is identically the same as seg- 
ment No. 1, of the diagram here presented. It may be perceived, 
also, that the half of segment No. 1, by its dotted transverse, is 
identically the same as one-fourth of segment No. 2, by its dotted 
transverse and longitudinal sections ; and that a right quarter sec- 
tion of No. 2 is identically the same as a right quarter section of 
No. 3, only reversed in position. 

Note. — The half of segment NTo. 2 or 3 by the dotted cross-section is a 
good working measure for the segments, and it may be taken out bearing the 
requisite margins for locks or laps, both along the curves and at the ward 
end. The allowance along the curve on one side should be with reference to 
the burr, and, on the other, to the ledge and lip. One of the pieces carrying 
the continuation for the pipe (oue of the outside pieces) should be fitted last, 
and trimmed to correct the angle, if necessary. But, with proper allowauce 
for locks, and correct handling in the edging-machine, trimming will seldom 
be required. 




PATTERN CUTTING. 



365 




The annexed diagram presents a side view of a 5-segment elbow, 
or, in other words, 
it shows the outline 
of the several seg- 
ments when they 
are rolled into place. 
d represents the di- 
ameter of the pipe, 
and r the radius of 
the throat. The 
dotted arcs have 
their centres at o. 
In this diagram, r is 
taken equal to fdf, 
and it should seldom 
or never be taken 
at less ; in practice 
it may be taken to 
any extent greater, 
a s circumstances 
may require. A 
3-piece elbow is by no means a handsome structure, but it is easily 
made. A 4-piece elbow looks well, though it has one of the angles 
of its perimeter vertical to the centre. 

Prob. 22. — To construct a Collar for a Cylindrical Pipe of the 
same Diameter as the Receiving-pipe. 
The unit measure of this description of collar has already been 
treated of, and the manner of constructing it explained. It is 
identically the same as that of a right-angled cylindrical elbow of 
the same diameter. Directions for its construction are given under 
Prob. 8. 

Prob. 23. — To apply the Unit Measure to the Construction of the 
Collar, Prob. 22. 

Rule. — Meas- 
ure down from the 
top of the plate 
from which the 
collar is to be tak- 
en, equal to the 
intended length of 
the collar and half 
its diameter, and 
draw a horizontal 

guide-line, B B B; next with the 
31* 




of the 



366 



PATTERN CUTTING. 



directly In the guide-line, scribe to its curve and its perpendicular 
in the several positions, 1, 2, 3, 4 ; then, B A B A B, with the 
continuation for the pipe above, will be the collar proposed. 

Note. — This collar will lock at one of its angles; and, ordinarily, this is 
the best practice. If it be desired to lock the collar from the crown of one of 
the arches, construct it by positions 2, 3, 4, and place position 1 in continuation 
of position 4. 

Peob. 24. — To construct a Cylindrical Collar of a given Diameter 
to Jit a Receiving-pipe or Cylinder of a greater given Diameter. 

Rule. — Construct a rectangle, A C B D, in length, A D, 

equal to ± part of the circum- 
ference of the collar ; and in 
breadth, A C, equal to £ the 
diameter of the collar multi- 
plied by the diameter of the 
collar and divided by the di- 
ameter of the receiving-pipe : 
make A e equal to the diam- 
eter of the collar multiplied 
by 0.5708. Proceed in all 
other respects for the unit 
measure by rule, Prob. 8. 

To apply the Unit Measure to the Construction of the 
Collar, Prob. 24. 




Prob. 25. 



Rule. — Proceed in all respects by rule, Prob. 23. 



By Mathematics. 



For the Unit Measure of a Right Cylindrical Collar of any given 
Diameter to Jit a Cylinder of a given Diameter. 

d — diameter of collar == conjugate axis of hyperbola. 
D == diameter of receiving-pipe or cylinder, 
C B = y = . 7854<2 = \ circ. of collar = base of measure. 
A C = x=d 2 -i-2D = perpendicular of measure. 
A e = i*d — d=.5708d. 

dg 

t = — j ($d-\- \? (jd)' 2 -\- y°~) = transverse axis of hyperbola. 

yt =^(te/ + ^); x i=y^dy + y»)-%t; x< being any 

abscissa or part of x, reckoned from the origin A, and y' the ordi- 
nate to abscissa x'. 



PATTERN CUTTING. 



367 



Prob. 26. — To construct a Cylindrical Collar to Jit an Elliptic- 
cylinder at either right section of the Ellipse. 

Rule. — Find, by any mechanical means, the radius of a circle 
arc that coincides best with the arc of the ellipse to be covered by 
the collar, and take twice that radius for the practical diameter, D, 
of the receiving pipe. Then proceed strictly by rule, Prob. 23, 
for the unit measure. Apply the measure to the construction of 
the collar by rule, Prob. 23. 

Note. — It is not in keeping with the laws of geometry to suppose that an 
ellipse and a circle can have a portion of their perimeters common to both ; 
but it is customary with sheet-iron workers to fix a cylindrical collar to an 
elliptic stove ; and by proceeding by the foregoing rule, a collar will be obtained 
which will practically fit without trimming, unless the diameter of the collar 
is unusually large for the size of the stove. 

Remark. — Section B A B o B of diagram, Prob. 22, shows the 
opening for the reception of the collar (or to be circumscribed by the 
collar) as it appears upon a plane surface. It may be formed upon 
a plane surface by scribing to the respective unit measure in the 
four positions indicated, in all cases. 

Prob. 27. — To construct a Cylindrical Collar of a given Diameter, 
to ft a Cylinder of the same Diameter, at any given Angle to the 
side of the Cylinder. 

d= diameter of collar, V wangle of collar to side of cylinder. 
Rule. — Construct a rectangle, A C B D, making A D equal 
to half the 
circumfe r- 
ence of the 
collar, and 
A C equal to 
£e?-htan£F. 
'Bisect 4 Z> in 
E and drop 
the perpen- 
dicular E JP, 
which will di- 
vide the rect- 
angle A C 
B D into two 
equal and 
similar rect- 
angles. Make 
B n equal to 
£c?X tan^F, 
and draw the line n o parallel to B F. Next, make A e and n e 




368 PATTERN CUTTING. 

each equal to 0.57086?, and proceed for the curves in all respects 
by rule, Prob. 8 : thus, A F n D will be the unit measure of the 
collar, and with the requisite continuation of the plate above A D, 
for the pipe, together with the necessary margins for the rivets and 
straight lock, will be equal to one-half of it. 

Example. — A cylindrical collar of \\ inches in diameter is to 
be constructed to fit a cylinder of the same diameter at an angle 
of 35° to the side of the cylinder ; then 

A Z>, diagram, = nd -J- 2 = 3.1416 X 4.5 -^- 2 = 7.0G86 inches ; 
and, by the table of natural sines, cosines, and tangents of different 
angles, it is found that the tangent of half the angle of 35° 
(tan 17° 30') is .31530 ; therefore, 

A C, diagram, = £rf-7-tan£F==2.25-f-.3153== 7.13606 in., and 
B n, diagram, = ±d X tan %V= 2.25 X .3153= .709425 in. 



Pbob. 28. — To construct a Cylindrical Collar, or Spout, of a given 
Diameter, to Jit a Cylinder of a greater given Diameter at a 
given Angle to the side of the Cylinder. 

d= diameter of collar. 
D = diameter of cylinder. 
V= proposed angle. 

Rule. — Make A D equal to 7r<2 2 -f- 2D, and in all other respects 
proceed strictly by rule, Prob. 27. 

Note. — Diagram, Prob. 27, represents a right semi-section of a collar, to iit 
a cylinder of the same diameter as the collar, at an angle of 45° to the side of 
the cylinder. It is a general guide for oblique collars. 



PATTERN CUTTING. 



369 



OF SPOUTS. 



Spouts for vessels are usually made conical, more or less, 
according to the stake on which they are turned ; and, by common 
consent, are divided into 
two classes ; viz., " tea- 
kettle spouts" to fit cylin- 
drical vessels ; and " cof- 
feepot spouts '" or " teapot 
spouts" to fit flaring ves- 
sels. 

In practice, no definite 
geometrical relations be- 
tween the spout and the 
vessel it is intended for 
are sought to be main- 
tained. Thus, the diam- 
eter of the spout relative 
to that of the body, its 
length, flare, angle of 
inclination to the body, 
and place of attachment, 
are matters of taste or 
convenience, or both, 
with the workman who 
constructs them. Never- 
theless, the ideas of sym- 
etry and" practical utility 
are not to be outraged, 
but, on the contrary, 
should be kept in view. 

Rules, strictly geomet- 
rical, might be given, 
covering probable cases ; 
but the workman, with a 
little practice, can much 
sooner design a becoming spout, and fit it satisfactorily by " trial 
and trimming," than in any other way. 

The annexed diagram represents the general outline of spouts. 
It is given in comparison with the true arm of a right cylindrical 
elbow, which it in a considerable degree resembles, that it may the 
more readily be understood. The full-lined figure is that of the 
spout ; that by the dotted lines, the arm of the elbow. 




370 



PATTERN CUTTING. 



Op Pitched or Bevelled Covers. 



Prob. 29. — To construct a Bevelled Circle, or Circular Cover, of 
a given Rise and given Diameter. 

Ti == rise, or perpendicular height. 
d = initial diameter, or diameter of hoop. 
s — slant height, or radius of chief arc, = o A , diagram. 
wi = sum of the widths of the burr and edge (about £ inch), 

= i£, diagram. 
s = \/((^dy -{- h 2 ) ; R = s -\- m, = o B, diagram. 

Kule. — From a common centre, o, properly chosen on the 
plate, describe two circles, taking R for the radius of one of them, 

and s for that of the other. 
Next, draw a radius from the 
centre to the outer circle, as 
o B, diagram, and cut out that 
circle. Next, cut a narrow 
flexible measure, in length 
equal to n(2R-d), and from 
the point B, with that measure 
bent to the circumference, 
measure off its length on the 
circumference, as from B to F, 
and from the new-found point, 
F, draw a radius to the centre, 
as o F. Next, bisect the arc 
B F, and draw the bisecting 
radius, o n. Next, allow the 
requisite margins for laps, 
seam, or lock, as by the dotted lines, and cut out by those lines. 
Lastly, cut open the remaining portion of the bisecting radius, to 
the centre, o : then will the sector, B D F o B, be the cover ; and 
the arc, ACE, will be equal to the circumference of the rim, or 
hoop. 

Example. — A bevelled cover is to be constructed having a rise 
of 1^ inches, and to fit a cylinder of 12 inches in diameter : then 

s = y/CC 1 !) 2 + 1.25 2 ] = 6.1289 inches ; 
R = 6.1289 -f- \ = 6.3789 inches ; and 

B n F, the length of the flexible measure, 
= 3.1416(6.3789 X 2 -12) = 2.381 inches. 

Note. — The value of s in all cases coming under this problem, since it is 
the length of the hypotenuse of a right-angled triangle whose base and per- 
pendicular are given, may be found by mechanical means, thus : construct a 




PATTERN CUTTING. 371 

right-angle by draft on a plane, making one of the legs equal to half the 
initial or given diameter (in the last example 6 inches), and the other equal 
to the given rise or perpendicular of the triangle (in the last example 
1| inches), then will the rectilinear distance between the extremities of the 
legs, opposite the angle, be the hypotenuse, or value of s, required; thus, 
either side of a right-angled triangle may be found, by mechanics, the other 
two sides being known. But I must not be understood by this as encouraging 
a desire to avoid the extraction of the square root of numbers by arithmetic ; 
without the ability to do that, the student will find himself often perplexed, 
and occasionally defeated. 

Prob. 30. — To construct a Pattern for a Bevelled Elliptic Cover 
of a given Rise, to ft an Elliptic Boiler of given Diameters. 

D = transverse diameter of boiler. 
d zzz conjugate diameter of boiler. 
h = rise or perpendicular height of cover. 

Kule.— Construct a right quarter-section of an ellipse, A C B, by- 
rule, Prob. 15 or 18, making A B equal to ^[(i^)) 2 4~ h 2 ], and B C 
equal to ^[(%df-\-h % ~\. 
MakeCn — l(P-p% 
P being the circum- 
ference of an ellipse 
whose semi-axes are 
V / [(^) 2 + ^ 2 ] and 

m¥) 2 + h *l and i> 
the circumference of 

one whose diameters 

are D and d; and 

from the point n draw 

a line n B : then will 

A n B be the unit 

measure of the cover, 

and will contain one-fourth part of it. Allow, as by the dotted 

lines, the necessary margins for the edge and half-lock. 

A B and B C may be found by mechanics (see Note appended 
to Prob. 29). 

Prob. 31. — To construct a Bevelled Cover of a given Rise, to fit a 
False-oval Boiler of given length and width. 

D= length or transverse diameter of boiler. 
d= width or conjugate diameter of boiler. 
h == rise or perpendicular height of cover. 
S= \/\_(h D r-\- h 2 ~\ = B E, diagram. 
N=(S Xd)-±-D = B m, diagram. 
h' = (dXh)-+D. 
F = \tl(±dy + h' 2 l = mn } diagram. 




372 



PATTERN CUTTING. 



\/(%D)* -j- h 2 , and m 



Rule. — Construct a rectangle, A B C D, making A D equal 
to £Z>, and A B and B m each equ al to \d. M ake B E equal to 

equal to y/(^) 2 + ^ 2 - Next ; with the 
square in position 
E H n, one of the 
blades cutting the 
points E and D, and 
the other cutting 
the point n, draw 
the lines E H, H n. 
Next, with the di- 
viders, find a radius, 
o B, that will cut 
the points B and n, 
and with o the cen- 
tre, describe the 
arc B n: then will B n H E be the unit measure of the cover, 
and contain one-fourth part of it, less the allowance, as by the dot- 
ted lines, for the edge and half the lock. 




Note. — When d = \D, h' = %h, and F, N, and B o = %S. S and F may be 
found by Mechanics, as by rule given in Note appended to Prob. 29. In prac- 
tice, if A be taken equal to j&P, the rise will generally be sufficient. 



Of Can-Tops. 

Can-Tops are simply truncated cones, and the cones themselves 
are pitched or bevelled circles. They may be defined in part by 
their pitch, which I shall here define to be the angle of the side of 
the cone to the base ; or they may be defined by their bases and 
perpendicular height. The body of a common tunnel is a two-thirds 
pitched can-top, or a can-top having a pitch of 60° ; or, in other 
words, it is the frustum of a cone, or pitched circle, whose slant 
height was equal to the diameter of the base : it is therefore made 
up of a semi-circle whose radius is equal to the greater base ; but 
can-tops are rarely pitched as steep as 60°. They may be con- 
structed in a single piece, and should be when practicable ; or they 
may be composed of two or more right-sections, as the body of a 
common flaring vessel ; so they may be pieced transversely, when 
desirable, after the manner of piecing a large tunnel. 



Prob. 32. — To construct a Can-top of a given Depth and given 
Diameters. 

Rule. — Proceed in all respects by rule, Prob. 1 or Prob. 2. 



PATTERN CUTTING. 



373 



Pros. 33. — To construct a Can-top of a given Pitch and given 
Diameters. 

D = diameter of greater base. 

d — diameter of lesser base. 
V= pitch, or angle of the side to the base. 
H= perpendicular height of generating cone, having D for its 

base. 
R = slant height of generating cone, having D for its base. 

r = slant height of cone, having d for its base. 






* = 2D— 



d)_ H(D-d) __ 
~ D 
height of the frustum. 



z^H-^-^dy, the 



Rule. — From a common centre describe two concentric arcs 
of circles of sufficient length, 
taking R for the radius of one 
of them, and r for that of the 
other. Next, draw a radius 
from the centre to the outer arc, 
as o r R, diagram. Next, with 
a flexible measure, cut to the re- 
quired length, (the whole cir- 
cumference of one of the bases, 
when practicable), and bent to 
the proper curve, measure off 
that length on the curve, as from 
R to v, and from the new-found 
point, v, draw a radius to the 
centre, o : then will R v n r be 
the top, or a known aliquot part of the top, required. 




Special Cases. 

For a two-thirds pitch, or pitch of 60° (sometimes erroneously 
called a half-pitch, because the angle of the side and axis is half as 
32 



374 PATTEKN CUTTING. 

great as that of the side and base) — From a common centre de- 
scribe two concentric semi-circles, taking the diameters of the given 
bases, respectively, for the radii : take the greater semi-circle for 
the top required. 

For a half-pitch, or pitch of 45°. — Describe two concentric 
circles, taking the diameters of the bases multiplied by 1.4142, 
respectively, for the radii ; take three-fourths of the greater circle 
for the body, less one-fourth of its radius as a chord. 

For a third-pitch, or pitch of 30 c (often erroneously called a half- 
pitch, because the angle of the base and side is half as great as 
that of the side and axis). — From a common centre describe two 
circles, taking D X 1.1549 for the radius of one, and d X 1.1549 
for that of the other. Next, draw a radius from the centre to the 
outer circumference. Next, take 3D — n =z Z)-f- 1.0472 in the di- 
viders, and from the summit of the radius just drawn, space it off 
on the outer circumference, as a chord ; and from the last found 
point draw a radius to the centre : the greater sector, or sector of 
the re-entrant angle, will be the body or top required. 

For a quarter-pitch, or pitch of 22^°. — From a common centre 
describe two circles, taking 1.0824Z) for the radius of one, and 
1.0824c? for that of the other. Next, draw a common radius to the 
circles. Next, take ^D in the dividers, and from the point where 
the radius cuts the circumference of the greater circle, step it off 
on that circumference as a chord ; and from the last found point 
draw a radius to the centre : take out the lesser sector, and the re- 
mainder of the circle will be the top required. 

Note 1. — The foregoing special cases are all covered, of course, by the 
preceding general rule ; but it is more expeditious to measure an arc by its 
chord, when the latter is known or can be readily found, than to measure it 
by the flexible measure, more generally employed. 

2. — Having laid off the dimensions pattern, as above directed, allow the 
requisite margins for burr and lock, or seam, as in other like cases. 

3. — A fifth-pitch, or pitch of 18°, is sometimes, but incorrectly, called a quar- 
ter-pitch, because the angle of the base and side is one-fourth as great as that 
of the axis and side. 

4. — The prevailing tendency of a can-top, or frustum of a cone, to " lop at 
the lock," as it is called, is clearly due to the careless manner of turning the 
wards and closing them. If the workman will turn the wards parallel to 
the lines of the dimensions pattern, and will use neither more nor less for the 
lock than the margins he allows, the tendency complained of will not obtain 
(see Note, Prob. 1). 

Of Lips for Measures. 

Lips for measures, when laid off on a plane, are simply lunes, or 
crescents. In practice, no arbitrary relations between the lip and 
the vessel it is intended for are sought to be maintained. Thus, 



PATTERN CUTTING. 375 

the length of the lip, its width, and its angle with the side of the 
vessel, are matters of taste or choice with the workman, limited, of 
course, by the purpose the lip is intended to subserve. 

The following rule, Prob. 34, which sets the lip at an angle of 
41° 50' to the side, very nearly, is the one most commonly in prac- 
tice ; and it appears to be as good a general rule as can be offered. 

Prob. 34. — To construct a Lip for a Measure, the Diameter of 
the Top of the Measure being given. 

Rule. — Take three-fourths of the diameter of the top of the 
measure in the dividers, and with that as radius describe a circle. 
Next, with the same radius, from a new centre, taken about mid- 
way between the centre of the circle and the circumference (more 
or less, according to the desired width of the lip, wired-edge includ- 
ed), describe an arc cutting both sides of the circumference : then 
will the crescent thus formed be the lip intended. 

Note.— To diminish the pitch, which serves to make the lip longer, take 
the radius at a greater ratio to the diameter of the lesser base, or top, than 
3 to 4. But, even when the measure has little or no flare, the radius should 
not much, if any, exceed seven-eighths of the diameter of the top. 



Of Sheet Pans. 

Sheet pans are vessels intended to hold fluids at a temperature 
above the melting point of solder. They are constructed of a sin- 
gle rectangular sheet, by turning up the sides, and folding the 
surplus surfaces at the corners upon the sides. 

Dripping-pans and oaldng-pans are commonly constructed with 
sides slightly flaring, while evaporating-pans, most generally, have 
perpendicular sides. 

In geometry, those having oblique sides are called prismoids, or 
prismoidal vessels, and those having perpendicular sides are called 
parallelopiped vessels, or prisms. 

These vessels are commonly constructed with wired tops, or rims, 
partly for the purpose of stiffening them, and partly to hold the 
sides in place. 

Sheet pans may be constructed to given dimensions, — length, 
breadth, and depth, — and thus to given capacities ; so they may bo 
constructed to given ratios of parts ; but, generally, economy of 
stock and utility of purpose, without further specifications, aro 
allowed to govern. 

With these remarks (and the additional one, perhaps, that the 
sides of a sheet pan are to be of equal width), we might dismisi 



376 



PATTERN CUTTING. 



the subject, were it not that rules for cutting the corners so as to 
bring the edges up to, or under, the wire, seem to be called for. 

Pros. 35.— To cut the Comers for a Perpendicular-sided Sheet Pan. 

Rule. — Let A B and D E, diagram, represent the width re- 
quired to cover the 
wire; B C and D C, 
the width of the 
sides. Remove the 
section A B m D E : 
then will the square 
BmD C, folded di- 
agonally, and reflex- 
ed upon the side or 
end, form the edge 
proposed. 




Prob. 36. — To cut the Corners for an Oblique-sided Sheet Pan. 

Rule. — Let A B and D E, diagram, represent the width or 
margins required to cover the wire ; B C and D C, the width of 
the sides ; B b and D d, the flare of the sides. Draw b n perpen- 
dicular to a C, and d n perpendicular to e C. Next, remove the 
section a b n d e: then will the trapezium b C d n, folded diago- 
nally from n to C, and reflexed upon one of the sides, form the 
edge contemplated. 

Prob. 37. — To construct a Heart, or Heart-shaped Cake-cutter. 

Rule. — Construct the Gothic arch, No. 1 ; this work, and upon 
its chord, describe two equal semi-circles, in diameter equal to half 
the length of the chord. 



Prob. 38. — To construct a Mouthpiece for a Speaking-tube. 

The curve for this structure consists of four cymas of unequal 
branches, in pairs. The structure itself is little else than two 
short, decidedly flaring teakettle spouts, in a single piece, having a 
side common to both. 



PATTERN CUTTING. 



377 



Pros. 39. — To construct a Pattern for the Body of a Circular- 
bottomed Flaring Coal-hod, all the curves to be arcs of circles. 

D = nominal diameter of greater base. 
d=. nominal diameter of lesser base. 
h = perpendicular depth of vessel. 

H=~; i2 = V / (0.61685D 2 +^); r==™- 

Rule. Place the square on the plate and scribe to its edges, as 
A t o, diagram ; produce A t in the direction B, and o t in the 
direction Z), making all 
the lines from t of suffi- 
cient length. Next, 
make t B equal to one- 
fourth part of the cir- 
cumference of the 
greater base =0. 78542), 
and make t A equal to 
t B. Next, make t o=H, 
— Dh-^-CD-d), and 
draw the radii o A, 
o B = R. Next, drop 
the square perpendicu- 
larly down on the line 
t o, equal to the given 
perpendicular depth, h, 
as from t to i, and 
through the point i draw 
the line m i n, parallel 
to the line A t B. Next, 
take the nominal diam- 
eter of the greater giv- 
en base (D) in the 
dividers, and with one 
foot in the point B, and 
the other in the line 
t o, as at /, and / the centre, describe the arc B D A. Next, with 
o A the radius and o the centre, describe the arc A C B. Next, 
with o m,z=:r, or o n the radius and o the centre, describe the arc 
m k n ; then will A D B n k m be the front half of the body, or a 
pattern covering that portion ; and A C B n k m, in a separate 
and distinct piece, will be the back half of the body, or a pattern 
covering that portion. Both portions are introduced here in the 

32* 




378 PATTERN CUTTING. 

same diagram, that their relations may be perceived and readily 
comprehended. 

Note. — This style of hod is introduced, partly to meet the popular demand 
with regard to the spout, partly with reference to economy in stock, and partly 
because of the readiness with which it may be plotted upon a plane surface, 
compared with the labor of plotting for an oval or elliptical bottom. The 
workman will find no difficulty in constructing it, except, perhaps, in turning 
the rim for the wire in the immediate vicinity of the side-locks : this he will 
probably be obliged to do on the stake. When locked at the sides and wired, 
it is to be compressed along the upper front half, and also in front, so as to 
form nearly a perpendicular- sided spout, of about four inches in width at 
the lip. 

The real bases of the vessel will be somewhat greater than the nominal, 
because the chords of the arcs are made equal to the given half-circumferences, 
instead of the arcs themselves ; the diameter of the bottom, therefore, will be 
equal to twice the arc m k n, divided by 3.1416, instead of being equal to d. 

In practice, if the nominal diameter of the greater base be taken equal to 
once and one-half that of the lesser, or the nominal diameter of the latter be 
taken equal to two-thirds that of the former, and the perpendicular depth be 
taken equal to the nominal diameter of the lesser base, the proportions will 
be found satisfactory ; moreover, if the nominal diameter of the greater base 
be taken at 12 inches, and the foregoing proportions maintained, a very fair 
medium-sized hod will be obtained, particularly if the requisite margins be 
allowed. 

The hoop for the bottom, which should be about one and a half inch in 
width after it is wired, may have the same flare as the body, and the radius of 
its lesser arc will be the same as that for the lesser base of the body. 



SOLDERS. — ALLOYS AND COMPOSITIONS. 379 



SOLDERS, ALLOYS, AND COMPOSITIONS. 

Hard solder. — Copper 2 parts, zinc 1 part ; — used with powdered borax. 

Pewterer's solder. — Tin 2 parts, antimony 1 part. 

Tinman's solder. — 1 part each, lead and tin. 

P lumber 's solder. — Tin 2 part, lead 5 parts ; or, pewter^ parts, tin 1, and bismuth 1 
Resin is used with the last three. 

Solder for iron. — Tough brass, used with borax. 

Silver solder. — 1 part brass, and from 2 to 5 parts fine silver. 

Spelter " for brass, copper, and German silver. — 2 parts brass, 1 part zinc. 

Solder for copper. — Brass 6 parts, tin 1, zinc 1. 

Dentist- s solder. — 4 parts 22 carat gold, 1 part silver, 1 part copper. 

Dentist's gold. — 10 parts 22 carat gold, 1 part silver, 1 part copper. 

Dentist's compound for clasps. — 5 parts 22 carat gold, 1 part platinum 

Yellow brass. — Copper 3 parts, zinc 1 part. 

Spelter. — Copper 2 parts, zinc 1 part. 

For lathe bushes- — Copper 16 parts, tin 4 parts, zinc 1 part. 
" " " harder. <— Copper 16 parts, tin 4 parts, zinc 2 parts. 

Improved Babbit metal. — This composition, for tbe lining of boxes, shaft bearings, &c, 
— which, from the satisfaction it has thus far given, bids fair to come into general use, — 
is composed of tin 12 parts, antimony 3 parts, and copper 2 parts. The original recipe 
for this alloy was, tin 6 parts, antimony regulus 2 parts, and copper 1 part, as its prime 
equivalents, to which, when about to be remelted for use, 2 parts of copper to 1 of the com- 
position were added. 

For pulley blocks. — Copper 7 parts, tin 1 part. 
" wheels, boxes, and cocks. — Copper 8 parts, tin 1 part. 

Bronze — government gun-metal. — 9 parts copper, 1 part tin. The specific gravity 
of this composition is greater than the mean of its constituents. 

For valves. — 10 parts copper, 1 part tin. 

Bell metal. — 39 parts copper, 11 parts tin. 

Gong metal — 40 parts copper, 5 tin, 2.8 zinc, 2.15 lead. 

Bath metal. — 32 parts brass, 9 parts zinc. 

Blanched copper. — 16 parts copper, 1 part arsenic. 

Britannia metal. — 1 part each, — brass, tin, bismuth, antimony. 

Petong, or Chinese white copper. — 20.2 parts copper, 15.8 nickel, 12.7 zinc, 1.3 iron. 

German silver. — 2 parts copper, 1 nickel, 1 zinc -, when intended to be rolled into plates, 
it is composed of 60 parts copper, 25 parts nickel, 20 of zinc, and 3 of lead. 

Manheim gold. — 3 parts copper, 1 of zinc, and a small quantity of tin. 

Mock gold. — 16 parts copper, 7 parts platinum, and 1 part zinc. 

Mock platinum. — 8 parts brass, 5 parts zinc. 

Speculum metal. — 7 parts copper, 3 zinc, 4 tin ; or, 6 parts copper, 2 of tin, and 1 of 
arsenic. 

Tombac, or gilding metal. — 9 parts copper, and 1 part zinc. 
Mock iron — expanding alloy. — Lead 9 parts, antimony 2 parts, bismuth 1 part. This 
composition expands in cooling, and is used in filling small defects in iron castings. 

Ring, or jeweller's gold. — 150 parts pure gold, 39 parts copper, 22 parts pure silver. 

Queen'* metal. — Tin 9 parts, antimony 1, lead 1, bismuth 1. 

Pewter, common. — Tin 4 parts, lead 1. 
" best. — Tin 100 parts, antimony 17. 

Steel alloyed with ^ part of platinum, or to the same extent with silver, is rendered 
harder, more malleable, and better adapted for every kind of cutting instrument. 



380 SOLDERS. — ALLOYS AND COMPOSITIONS. 

Solder for gold. — 3 parts gold, 1 part silver, 1 part copper. 

Solder for Britannia. — Tin, 7 parts ; lead, 4 parts. 

Yellow solder. — Copper, 1 part; zinc, 1 part. 

Black solder. — Copper and zinc, each 8 parts ; tin, 1 part. 

Pewterer's soft solder. — Bismuth, 1 part; tin, 2 parts ; lead, 1 part 

Common Britannia metal. — Tin, 100 parts ; copper, 2 parts ; antimony, 1 part. 

Common bronze metal. — Copper, 4 parts ; zinc, 2 parts ; tin, 1 part. 

White metal. — 5 parts copper, 3 zinc, 1 lead, 1 tin. 

Silver-colored metal. — T^u, 50 parts ; copper, 3 parts ; antimony, 3 parts ; bis- 
muth, 1 part. 

Imitation silver. — 16 parts copper, 1 part zinc. 

Pinchbeck. — Copper, 4 parts ; zinc 2 1 part. 

Metal for taking Impressions. — Bismuth, 6 parts; lead, 2 parts ; tin, lpart. 

Rivet metal. — Copper, 10 parts ; tin, 5 parts ; zinc, 2 parts. 

Fusible alloy (melts at 200°). — Bismuth, 2 parts ; lead, 1 part ; tin, 1 part. 

Muriate of zinc. — Muriatic acid, holding in solution all the zinc it will dissolve 

Acid for soldering tin. — Muriate of zinc, 1 part by volume; soit water, 2 parts 
by volume ; add a trifle of Sal. Ammoniac. 

Acid for soldering zinc. — Muriate of zinc, 10 ounces ; Sal. Ammoniac, 1 ounce ; 
water, 1 pint. 

Acid for soldering brass or copper. — Muriate of zinc, 5 parts ; Sal. Ammoniac, 1 
part. 

Acid for soldering gold or silver. — Muriatic Acid, 2 parts ; sperm tallow, 1 part ; 
Sal. Ammoniac, 1 part; all by weight. 

Arid for soldering iron. — Muriatic Acid, 16 parts; sperm tallow, 6 parts; Sal. 
Ammoniac, 4 parts ; all by weight. 

Tinning acid, for brass or copper —Muriate of zinc, 4 parts ; soft water, 4 parte ; 
Sal. Ammoniac, 1 part ; all by weight. 



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